#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define pi pair<int,int>
#define vi vector<int>
#define rep(i,x,n) for(int i=x; i<n; ++i)
#define For(i,n) rep(i,0,n)
#define endl "\n"
#define sp ' '
#define pb push_back
#define f first
#define s second
#define sz size()
#define all(x) (x).begin(),(x).end()
const int N = 20, OO = 2e9, mod = 1e9+7, mx = 1e3+1;
const int dx[]{0,1}, dy[]{1,0};
void tr(int a, int b){cout << a << sp << b << endl;}
void cmx(int &a, int b){a = max(a,b);}
void cmn(int &a, int b){a = min(a,b);}
int dp[1<<N], a[N], b[N];
bitset <mx> bt[1<<N];
int32_t main() {
ios::sync_with_stdio(0); cin.tie(0);
// freopen("talent.in", "r", stdin);
// freopen("talent.out", "w", stdout);
int n,m; cin >> n >> m;
For(i,n) cin >> a[i];
For(i,m) cin >> b[i];
For(i,m){
bt[1<<i][0] = 1;
bt[1<<i] |= (bt[1<<i] << b[i]);
if(bt[1<<i][a[0]]){
dp[1<<i]++;
}
}
for(int msk=3; msk < 1<<m; ++msk){
if(__builtin_popcount(msk)==1) continue;
For(i,m) if((1<<i)&msk){
int other = msk - (1<<i);
bitset<mx> cur = bt[other];
cur |= (cur<<b[i]);
if(dp[other] < n && cur[a[dp[other]]]){
if(dp[msk] < dp[other]+1){
dp[msk] = dp[other]+1;
bt[msk].reset();
bt[msk][0] = 1;
}
}else{
if(dp[msk] < dp[other]){
dp[msk] = dp[other];
bt[msk] = cur;
} else if(dp[msk]==dp[other]){
bt[msk] |= cur;
}
}
}
}
if(dp[(1<<m)-1] == n) cout << "YES";
else cout << "NO";
cout << endl;
return 0;
}
/*
Just some notices :
I believe you can do it !
You've done things that were harder ...
Stay calm and focused =)
*/
