#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define range(a) a.begin(), a.end()
#define endl "\n"
#define Yes() cout << "Yes" << endl
#define No() cout << "No" << endl
#define MP make_pair
using Graph = vector<vector<int>>;
const unsigned long long mod = 1e9 + 7;
const long long INF = 1LL<<60;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
void chmin(long long &a, long long b) { if (a > b) a = b; }
void chmax(long long &a, long long b) { if (a < b) a = b; }

//(xi=f)∨(xj=g)というクローズを足し、これをすべて満たす変数の割当があるかを解きます。(ACL pre H参照)
namespace internal {

template <class E> struct csr {
    vector<int> start;
    vector<E> elist;
    csr(int n, const vector<pair<int, E>>& edges)
        : start(n + 1), elist(edges.size()) {
        for (auto e : edges) {
            start[e.first + 1]++;
        }
        for (int i = 1; i <= n; i++) {
            start[i] += start[i - 1];
        }
        auto counter = start;
        for (auto e : edges) {
            elist[counter[e.first]++] = e.second;
        }
    }
};

// Reference:
// R. Tarjan,
// Depth-First Search and Linear Graph Algorithms
struct scc_graph {
  public:
    scc_graph(int n) : _n(n) {}

    int num_vertices() { return _n; }

    void add_edge(int from, int to) { edges.push_back({from, {to}}); }

    // @return pair of (# of scc, scc id)
    pair<int, vector<int>> scc_ids() {
        auto g = csr<edge>(_n, edges);
        int now_ord = 0, group_num = 0;
        vector<int> visited, low(_n), ord(_n, -1), ids(_n);
        visited.reserve(_n);
        auto dfs = [&](auto self, int v) -> void {
            low[v] = ord[v] = now_ord++;
            visited.push_back(v);
            for (int i = g.start[v]; i < g.start[v + 1]; i++) {
                auto to = g.elist[i].to;
                if (ord[to] == -1) {
                    self(self, to);
                    low[v] = min(low[v], low[to]);
                } else {
                    low[v] = min(low[v], ord[to]);
                }
            }
            if (low[v] == ord[v]) {
                while (true) {
                    int u = visited.back();
                    visited.pop_back();
                    ord[u] = _n;
                    ids[u] = group_num;
                    if (u == v) break;
                }
                group_num++;
            }
        };
        for (int i = 0; i < _n; i++) {
            if (ord[i] == -1) dfs(dfs, i);
        }
        for (auto& x : ids) {
            x = group_num - 1 - x;
        }
        return {group_num, ids};
    }

    vector<vector<int>> scc() {
        auto ids = scc_ids();
        int group_num = ids.first;
        vector<int> counts(group_num);
        for (auto x : ids.second) counts[x]++;
        vector<vector<int>> groups(ids.first);
        for (int i = 0; i < group_num; i++) {
            groups[i].reserve(counts[i]);
        }
        for (int i = 0; i < _n; i++) {
            groups[ids.second[i]].push_back(i);
        }
        return groups;
    }

  private:
    int _n;
    struct edge {
        int to;
    };
    vector<pair<int, edge>> edges;
};

}  // namespace internal

// Reference:
// B. Aspvall, M. Plass, and R. Tarjan,
// A Linear-Time Algorithm for Testing the Truth of Certain Quantified Boolean
// Formulas
struct two_sat {
  public:
    two_sat() : _n(0), scc(0) {}
    two_sat(int n) : _n(n), _answer(n), scc(2 * n) {}

    void add_clause(int i, bool f, int j, bool g) {
        assert(0 <= i && i < _n);
        assert(0 <= j && j < _n);
        scc.add_edge(2 * i + (f ? 0 : 1), 2 * j + (g ? 1 : 0));
        scc.add_edge(2 * j + (g ? 0 : 1), 2 * i + (f ? 1 : 0));
    }
    bool satisfiable() {    //条件を足す割当が存在するかどうかを判定する
        auto id = scc.scc_ids().second;
        for (int i = 0; i < _n; i++) {
            if (id[2 * i] == id[2 * i + 1]) return false;
            _answer[i] = id[2 * i] < id[2 * i + 1];
        }
        return true;
    }
    vector<bool> answer() { return _answer; }   //最後に呼んだ satisfiable の、クローズを満たす割当を返す。

  private:
    int _n;
    vector<bool> _answer;
    internal::scc_graph scc;
};


int main(void){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int N, M;
	cin >> N >> M;
	vector<int> L(N), R(N), NL(N), NR(N);
	FOR(i,0,N){
		ll a, b;
		cin >> a >> b;
		L.at(i) = a;
		R.at(i) = b;
		NR.at(i) = M - L.at(i) - 1;
		NL.at(i) = M - R.at(i) - 1;
	}

	two_sat tf(N);
	FOR(i,0,N-1){
		FOR(j,i+1,N){
			if(L.at(i)<=R.at(j)&&R.at(i)>=L.at(j)){
				tf.add_clause(i, false, j, false);
			}
			if(L.at(i)<=NR.at(j)&&R.at(i)>=NL.at(j)){
				tf.add_clause(i, false, j, true);
			}
			if(NL.at(i)<=R.at(j)&&NR.at(i)>=L.at(j)){
				tf.add_clause(i, true, j, false);
			}
			if(NL.at(i)<=NR.at(j)&&NR.at(i)>=NL.at(j)){
				tf.add_clause(i, true, j, true);
			}
		}
	}

	if(tf.satisfiable())
		cout << "Yes" << endl;
	else
		cout << "No" << endl;

	return 0;
}