#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll=long long;
using ld=long double;
ld pie=3.141592653589793;
ll mod=1000000007;
ll inf=9999999999999999;
int main(){
    ll n,k;
    cin >> n >> k;
    vector<ll>a(n);
    for (ll i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    sort(a.begin(),a.end());
    vector<ll>thr(21,1);
    for (ll i = 1; i < 21; i++)
    {
        thr[i]=thr[i-1]*3;
    }
    if (n>10)
    {
        map<ll,ll>memo1;
        ll d=n-10;
        for (ll i = 0; i < thr[10]; i++)
        {
            ll now=0;
            vector<ll>z(3,0);
            for (ll j = 0; j <10 ; j++)
            {
                ll x=i%thr[j+1];
                x/=thr[j];
                if (x==0)
                {
                    now+=a[j];
                }else if (x==1)
                {
                    now-=a[j];
                }
                z[x]+=1;
            }
            if (z[0]==10||z[1]==10||z[2]==10)
            {
                continue;
            }
            memo1[now]+=1;
        }
        map<ll,ll>memo2;
        for (ll i = 0; i < thr[d]; i++)
        {
            ll now=0;
            vector<ll>z(3,0);
            for (ll j = 0; j <d ; j++)
            {
                ll x=i%thr[j+1];
                x/=thr[j];
                if (x==0)
                {
                    now+=a[j+10];
                }else if (x==1)
                {
                    now-=a[j+10];
                }
                z[x]+=1;
            }
            if (z[0]==d||z[1]==d||z[2]==d)
            {
                continue;
            }
            memo2[now]+=1;
        }
        for(auto v:memo1){
            if (v.first==k)
            {
                cout << "Yes" << endl;
                return 0;
            }
            if (memo2[k+v.first]>=1||memo2[v.first-k]>=1)
            {
                cout << "Yes" << endl;
                return 0;
            }
        }
        cout << "No" << endl;
    }else{
        map<ll,ll>memo1;
        for (ll i = 0; i < thr[n]; i++)
        {
            ll now=0;
            vector<ll>z(3,0);
            for (ll j = 0; j <n ; j++)
            {
                ll x=i%thr[j+1];
                x/=thr[j];
                if (x==0)
                {
                    now+=a[j];
                }else if (x==1)
                {
                    now-=a[j];
                }
                z[x]+=1;
            }
            if (z[0]==n||z[1]==n||z[2]==n)
            {
                continue;
            }
            memo1[now]+=1;
        }
        if (memo1[k]>=1)
        {
            cout << "Yes" << endl;
        }else{
            cout << "No" << endl;
        }
        
    }
    
}