#include <bits/stdc++.h>
using namespace std;
constexpr int mxN = 20, mxSum = 1000;
int n, m;
int a[mxN], b[mxN];
bool dp[1<<20];
int main(){
cin.tie(nullptr)->sync_with_stdio(false);
cin >> n >> m;
for (int i=0; i<n; ++i)
cin >> a[i];
for (int i=0; i<m; ++i)
cin >> b[i];
dp[0] = true; // Base case: Empty set can form sum 0
for (int mask=0; mask < (1<<m); ++mask){
if (!dp[mask]) continue;
int sum = 0;
for (int i=0; i<m; ++i)
if (mask & (1<<i)) sum += b[i];
for (int i=0; i<m; ++i){
if (!(mask & (1<<i)) && sum + b[i] <= mxSum)
dp[mask | (1<<i)] = true;
}
}
// Check if we can form each salary exactly
for (int i=0; i<n; ++i){
bool canPay = false;
for (int mask=0; mask < (1<<m); ++mask){
int sum = 0;
for (int j=0; j<m; ++j)
if (mask & (1<<j)) sum += b[j];
if (sum == a[i]){
canPay = true;
break;
}
}
if (!canPay){
cout << "NO\n";
return 0;
}
}
cout << "YES\n";
}
