#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ff(i,a,b) for(int i=a;i<=b;i++)
#define fb(i,b,a) for(int i=b;i>=a;i--)
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 21;
const int inf = 1e9 + 5;
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k) print the k-th smallest number in os(0-based)
int n, m;
int A[maxn];
int B[maxn];
bool dp[20][(1 << 20)];
int main()
{
ios::sync_with_stdio(false);
cout.tie(nullptr);
cin.tie(nullptr);
cin >> n >> m;
ff(i,0,n - 1)cin >> A[i];
ff(i,0,m - 1)cin >> B[i];
dp[0][0] = 1;
ff(i,0,n - 1){
ff(mask,0,(1 << m) - 1){
// ako za prve i koristim mask, da ih dobijem sve
// sad treba da vidim koje sve submaske mi daju zbir jednak A[i]
int smask = mask;
while(smask > 0){
int zbir = 0;
ff(j,0,m - 1){
if(smask & (1 << j))zbir += B[j];
}
if(zbir == A[i])dp[i + 1][mask] |= dp[i][mask ^ smask];
smask = (smask - 1) & mask;
}
}
}
ff(mask,0,(1 << m) - 1){
if(dp[n][mask] == 1)return cout << "YES", 0;
}
cout << "NO" << endl;
return 0;
}
/**
// probati bojenje sahovski ili slicno
**/
