#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
// constexpr int MOD = 998244353;
constexpr int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
constexpr int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U> inline bool chmax(T &a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U> inline bool chmin(T &a, U b) { return a > b ? (a = b, true) : false; }
struct IOSetup {
  IOSetup() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << fixed << setprecision(20);
  }
} iosetup;

bool solve(vector<int> &a, ll k) {
  if (k == 0) return true;
  int n = a.size();
  if (n == 0) return false;
  if (n == 1) return k == a[0] && k == -a[0];
  vector<ll> l{0};
  REP(i, n / 2) {
    vector<ll> nx(l);
    for (ll e : l) {
      nx.emplace_back(e - a[i]);
      nx.emplace_back(e + a[i]);
    }
    l.swap(nx);
    sort(ALL(l));
    l.erase(unique(ALL(l)), l.end());
  }
  vector<ll> r{0};
  FOR(i, n / 2, n) {
    vector<ll> nx(r);
    for (ll e : r) {
      nx.emplace_back(e - a[i]);
      nx.emplace_back(e + a[i]);
    }
    r.swap(nx);
    sort(ALL(r));
    r.erase(unique(ALL(r)), r.end());
  }
  for (ll e : l) {
    auto it = lower_bound(ALL(r), k - e);
    if (it != r.end() && e + *it == k) return true;
  }
  return false;
}

int main() {
  int n; ll k; cin >> n >> k;
  vector<int> a(n); REP(i, n) cin >> a[i];
  if (count(ALL(a), k) > 0) {
    cout << "Yes\n";
    return 0;
  }
  REP(i, n) REP(j, n) {
    if (i != j) {
      vector<int> b;
      REP(x, n) {
        if (x != i && x != j) b.emplace_back(a[x]);
      }
      if (solve(b, k - a[i] + a[j])) {
        cout << "Yes\n";
        return 0;
      }
    }
  }
  cout << "No\n";
  return 0;
}