#include <bits/stdc++.h>
#define int long long
#define ll long long
#define ld long double
#define ull unsigned long long
#define endl '\n'
#define NF string::npos
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define FAST ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
/*//! Arithmetic sequence law
(n * (2*a + (n-1)*d))/2;
n -> number of terms
a -> first term
d -> common difference
*/
using namespace std;
const int N = 1e5 + 5;
const int mod = 998244353;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int n, m;
vector <int> a, b;
vector <pair <int, int> > dp(1<<20, {-1, -1});
pair <int, int> rec(int mask)
{
if(__builtin_popcount(mask) == m) return {0, 0};
pair <int, int> &ret = dp[mask];
if(~ret.first) return ret;
ret = {0, 0};
for(int i = 0; i < m; i++)
{
if(mask & (1<<i)) continue;
pair <int, int> p = rec(mask | (1<<i));
if(ret.first == n) continue;
int idx = p.first, sum = p.second;
if(idx == n)
{
ret = p;
continue;
}
idx = p.first, sum = p.second + b[i];
if(sum == a[idx]) idx++, sum = 0;
ret = max(ret, {idx, sum});
}
return ret;
}
signed main()
{
FAST;
cin >> n >> m;
a.resize(n);
b.resize(m);
for(int i = 0; i < n; i++) cin >> a[i];
for(int i = 0; i < m; i++) cin >> b[i];
pair <int, int> ans = rec(0);
cout << (ans.first == n ? "YES" : "NO");
// cout << ans.first << " " << ans.second;
return 0;
}
