#ifndef HIDDEN_IN_VS // 折りたたみ用

// 警告の抑制
#define _CRT_SECURE_NO_WARNINGS

// ライブラリの読み込み
#include <bits/stdc++.h>
using namespace std;

// 型名の短縮
using ll = long long; // -2^63 ～ 2^63 = 9 * 10^18（int は -2^31 ～ 2^31 = 2 * 10^9）
using pii = pair<int, int>;	using pll = pair<ll, ll>;	using pil = pair<int, ll>;	using pli = pair<ll, int>;
using vi = vector<int>;		using vvi = vector<vi>;		using vvvi = vector<vvi>;
using vl = vector<ll>;		using vvl = vector<vl>;		using vvvl = vector<vvl>;
using vb = vector<bool>;	using vvb = vector<vb>;		using vvvb = vector<vvb>;
using vc = vector<char>;	using vvc = vector<vc>;		using vvvc = vector<vvc>;
using vd = vector<double>;	using vvd = vector<vd>;		using vvvd = vector<vvd>;
template <class T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
using Graph = vvi;

// 定数の定義
const double PI = acos(-1);
const vi DX = { 1, 0, -1, 0 }; // 4 近傍（下，右，上，左）
const vi DY = { 0, 1, 0, -1 };
int INF = 1001001001; ll INFL = 4004004004004004004LL;
double EPS = 1e-12;

// 入出力高速化
struct fast_io { fast_io() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(18); } } fastIOtmp;

// 汎用マクロの定義
#define all(a) (a).begin(), (a).end()
#define sz(x) ((int)(x).size())
#define lbpos(a, x) (int)distance((a).begin(), std::lower_bound(all(a), x))
#define ubpos(a, x) (int)distance((a).begin(), std::upper_bound(all(a), x))
#define Yes(b) {cout << ((b) ? "Yes\n" : "No\n");}
#define rep(i, n) for(int i = 0, i##_len = int(n); i < i##_len; ++i) // 0 から n-1 まで昇順
#define repi(i, s, t) for(int i = int(s), i##_end = int(t); i <= i##_end; ++i) // s から t まで昇順
#define repir(i, s, t) for(int i = int(s), i##_end = int(t); i >= i##_end; --i) // s から t まで降順
#define repe(v, a) for(const auto& v : (a)) // a の全要素（変更不可能）
#define repea(v, a) for(auto& v : (a)) // a の全要素（変更可能）
#define repb(set, d) for(int set = 0; set < (1 << int(d)); ++set) // d ビット全探索（昇順）
#define repp(a) sort(all(a)); for(bool a##_perm = true; a##_perm; a##_perm = next_permutation(all(a))) // a の順列全て（昇順）
#define smod(n, m) ((((n) % (m)) + (m)) % (m)) // 非負mod
#define uniq(a) {sort(all(a)); (a).erase(unique(all(a)), (a).end());} // 重複除去
#define EXIT(a) {cout << (a) << endl; exit(0);} // 強制終了

// 汎用関数の定義
template <class T> inline ll pow(T n, int k) { ll v = 1; rep(i, k) v *= n; return v; }
template <class T> inline bool chmax(T& M, const T& x) { if (M < x) { M = x; return true; } return false; } // 最大値を更新（更新されたら true を返す）
template <class T> inline bool chmin(T& m, const T& x) { if (m > x) { m = x; return true; } return false; } // 最小値を更新（更新されたら true を返す）

// 演算子オーバーロード
template <class T, class U> inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template <class T> inline istream& operator>>(istream& is, vector<T>& v) { repea(x, v) is >> x; return is; }
template <class T> inline vector<T>& operator--(vector<T>& v) { repea(x, v) --x; return v; }
template <class T> inline vector<T>& operator++(vector<T>& v) { repea(x, v) ++x; return v; }

// 手元環境（Visual Studio）
#ifdef _MSC_VER
#include "local.hpp"
// 提出用（gcc）
#else
inline int popcount(int n) { return __builtin_popcount(n); }
inline int popcount(ll n) { return __builtin_popcountll(n); }
inline int lsb(int n) { return n != 0 ? __builtin_ctz(n) : -1; }
inline int lsb(ll n) { return n != 0 ? __builtin_ctzll(n) : -1; }
inline int msb(int n) { return n != 0 ? (31 - __builtin_clz(n)) : -1; }
inline int msb(ll n) { return n != 0 ? (63 - __builtin_clzll(n)) : -1; }
#define gcd __gcd
#define dump(...)
#define dumpel(v)
#define dump_list(v)
#define input_from_file(f)
#define output_to_file(f)
#define Assert(b) { if (!(b)) while (1) cout << "OLE"; }
#endif

#endif // 折りたたみ用


//--------------AtCoder 専用--------------
#include <atcoder/all>
using namespace atcoder;

using mint = modint1000000007;
//using mint = modint998244353;
//using mint = modint; // mint::set_mod(m);

istream& operator>>(istream& is, mint& x) { ll x_; is >> x_; x = x_; return is; }
ostream& operator<<(ostream& os, const mint& x) { os << x.val(); return os; }
using vm = vector<mint>; using vvm = vector<vm>; using vvvm = vector<vvm>;
//----------------------------------------


//【局面のニム値】O(?)（遅いので実験用）
/*
* 初期局面 p から遷移可能な局面とそのニム値のリストを nim に格納する．
* nxt(p, nps) を呼ぶと，p から遷移可能な局面のリストを nps に格納するものとする．
*/
template <class T>
void calc_nimber(const T& p, function<void(const T&, vector<T>&)>& nxt, map<T, int>& nim) {
	nim.clear();

	function<int(const T&)> calc_nimber = [&](const T& p) {
		if (nim.count(p)) return nim[p];

		vector<T> nps;
		nxt(p, nps);

		vi next_nimbers;
		repe(np, nps) {
			next_nimbers.push_back(calc_nimber(np));
		}
		uniq(next_nimbers);

		int i = 0;
		while (i < sz(next_nimbers) && next_nimbers[i] == i) i++;
		nim[p] = i;

		return nim[p];
	};

	calc_nimber(p);
}


void zikken() {
	using T = vi;
	function<void(const T&, vector<T>&)> nxt = [&](const T& p, vector<T>& nps) {
		rep(i, sz(p)) {
			if (p[i] == 1) continue;

			if ((i == 0 || p[i - 1] == 1) && (i == sz(p) - 1 || p[i + 1] == 1)) {
				T np(p);
				np[i] = 1;
				nps.push_back(np);
			}
		}

		repi(i, 1, sz(p) - 2) {
			if (p[i - 1] == 1 || p[i] == 1 || p[i + 1] == 1) continue;

			T np1(p);
			np1[i] = np1[i - 1] = 1;
			nps.push_back(np1);

			T np2(p);
			np2[i] = np2[i + 1] = 1;
			nps.push_back(np2);
		}
	};

	repi(n, 0, 25) {
		T p(n + 1);
		p[0] = 1;

		map<T, int> nim;
		calc_nimber(p, nxt, nim);

		dump(n, nim[p]);
	}
	exit(0);

	repi(n, 0, 25) {
		T p(n + 2);
		p[0] = p[n + 1] = 1;

		map<T, int> nim;
		calc_nimber(p, nxt, nim);

		dump(n, nim[p]);
	}

	exit(0);
}
/*
片側の開いた長さ i の '0' の連
0 0
1 1
2 0
3 0
4 1
5 2
6 2
7 1
8 4
9 0
10 1
11 4
12 2
13 1
14 4
15 0
16 1
17 4
18 2
19 1
20 4
21 2
22 1
23 0
24 2
25 1

両側とも閉じた長さ i の '0' の連
0 0
1 1
2 0
3 0
4 1
5 2
6 2
7 1
8 4
9 0
10 1
11 4
12 2
13 1
14 4
15 0
16 1
17 4
18 2
19 1
20 4
21 2
22 1
23 0
24 2
25 1

http://oeis.org/A071430
Sprague-Grundy values for octal game .16
山から 1 つの石を取り除く場合，残る山の数を 0 にすることができる．
山から 2 つの石を取り除く場合，残る山の数を 1 または 2 にすることができる．

Sequence is eventually periodic with period 149459. The last exception is at n=105350.
*/


string TLE(int n, ll k, vl a) {
	a.insert(a.begin(), 0);
	a.push_back(k + 1);

	chmax(k, 10LL);

	vi dp(k + 1);
	dp[0] = 0;
	dp[1] = 1;
	dp[2] = 0;
	repi(i, 3, k) {
		int set = 0;
		repi(j, 0, (i - 2) / 2) {
			set |= 1 << (dp[j] ^ dp[i - j - 2]);
		}
		dp[i] = lsb(~set);
	}
	dump(dp);

	int res = 0;

	rep(i, n + 1) {
		ll w = a[i + 1] - a[i] - 1;
		dump(w);

		res ^= dp[w];
	}

	return res ? "Yes" : "No";
}


//【ランレングス符号】O(n)
/*
* a[0..n) をランレングス符号化し，結果を cls に格納する．
* cls[i] = {c, l} は前から i 番目の列が l 個の文字 c からなることを表す．
*/
template <class T>
void run_length_encodeing(const vector<T>& a, vector<pair<T, int>>& cls) {
	// verify : https://atcoder.jp/contests/arc024/tasks/arc024_2

	int n = sz(a);
	cls.clear();

	if (n == 0) return;

	cls.push_back({ a[0], 1 });

	// 今読んでいる文字の種類を記憶する．
	T c = a[0];

	repi(i, 1, n - 1) {
		// 記憶している文字と同じ文字の場合
		if (c == a[i]) {
			// 列の長さを増やす．
			cls.rbegin()->second++;
		}
		// 記憶している文字と異なる文字の場合
		else {
			// 新しい文字を記憶しておく．
			c = a[i];

			// 新たな列を追加する．
			cls.push_back({ c, 1 });
		}
	}
}


void umekomi() {
	auto start = chrono::system_clock::now();

	int k = 149459 + 105350 + 10;
	k = k / 3 * 3;
//	k = 1002; // デバッグ用

	vi dp(k);
	dp[0] = 0;
	dp[1] = 1;
	dp[2] = 0;
	repi(i, 3, k - 1) {
		int set = 0;
		repi(j, 0, (i - 2) / 2) {
			set |= 1 << (dp[j] ^ dp[i - j - 2]);
		}
		dp[i] = lsb(~set);
	}
//	dump(dp);
	
	rep(i, k) {
		if (14 <= dp[i] && dp[i] <= 17) dp[i] -= 4;
		else if (22 <= dp[i] && dp[i] <= 23) dp[i] -= 8;
	}

	vi dp3(k / 3);
	rep(i, k / 3) dp3[i] = ((dp[3 * i] * 16) + dp[3 * i + 1]) * 16 + dp[3 * i + 2];
//	dump(dp3);

	vector<pii> cls;
	run_length_encodeing(dp3, cls);
//	dump(cls);

	cout << "string s = R\"(";
	repe(cl, cls) {
		int c, l;
		tie(c, l) = cl;

		char c1 = ':' + c / 64;
		char c2 = ':' + c % 64;

		cout << c1 << c2;
		if (l >= 2) cout << l;
	}
	cout << "~)\";" << endl;

	auto now = chrono::system_clock::now();
	auto msec = chrono::duration_cast<std::chrono::milliseconds>(now - start).count();
	dump(msec);

	exit(0);
}
/*
24859 (25秒で終わった)
*/


const char* s = 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string solve(int n, ll k, vl a) {
	vi dp;

	int i = 0;
	while (s[i] != '~') {
		int c1 = s[i++] - ':';
		int c2 = s[i++] - ':';
		int c = c1 * 64 + c2;

		int l = 0;
		while (s[i] <= '9') l = l * 10 + s[i++] - '0';
		if (l == 0) l = 1;
		
		int d0 = c / 16 / 16;
		int d1 = c / 16 % 16;
		int d2 = c % 16;

		rep(hoge, l) {
			dp.emplace_back(d0);
			dp.emplace_back(d1);
			dp.emplace_back(d2);
		}
	}
	dump(dp);

	rep(i, sz(dp)) {
		if (14 - 4 <= dp[i] && dp[i] <= 17 - 4) dp[i] += 4;
		else if (22 - 8 <= dp[i] && dp[i] <= 23 - 8) dp[i] += 8;
	}
	dump(dp);

	a.insert(a.begin(), 0);
	a.push_back(k + 1); 
	
	int res = 0;

	rep(i, n + 1) {
		ll w = a[i + 1] - a[i] - 1;
		dump(w);

		if (w >= sz(dp)) {
			w = (w - 105350) % 149459 + 105350;
		}

		res ^= dp[w];
	}

	return res ? "Yes" : "No";
}


int main() {
//	input_from_file("input.txt");
//	output_to_file("output.txt");
	
//	umekomi();

	int n; ll k;
	cin >> n >> k;

	vl a(n);
	cin >> a;

	cout << solve(n, k, a) << endl;
}