#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename K> using hset = gp_hash_table<K, null_type>;
template <typename K, typename V> using hmap = gp_hash_table<K, V>;
using namespace std;
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define eb emplace_back
#define smax(x, y) (x = max(x, y))
#define smin(x, y) (x = min(x, y))
using ll = long long;
using ld = long double;
template <typename T>
using vv = vector<vector<T>>;
using vi = vector<int>;
using ii = array<int, 2>;
using vii = vector<ii>;
using vvi = vv<int>;
using vll = vector<ll>;
using l2 = array<ll, 2>;
using vl2 = vector<l2>;
using vvll = vv<ll>;
template <typename T>
using minq = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxq = priority_queue<T>;
const ll M = 1000000007;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, m;
cin >> n >> m;
// given subset of notes, whats the largest prefix we can satisfy?
// we're going to have a prefix with some leftover
// consider removing a note to get some (i, l)
// if
vi a(n), b(m);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
for (int i = 0; i < m; ++i) {
cin >> b[i];
}
// dp(s) = max dp(s\{b[i]}) + thing
vii dp(1 << m, {-M, -M});
dp[0] = {0, 0};
for (int s = 1; s < (1 << m); ++s) {
for (int i = 0; i < m; ++i) {
if (s & (1 << i)) {
auto [p, l] = dp[s ^ (1 << i)];
int np = p, nl = l + b[i];
if (nl == a[p]) {
np++;
nl = 0;
}
ii ndp = {np, nl};
smax(dp[s], ndp);
}
}
if (dp[s][0] == n) {
cout << "YES\n";
return 0;
}
}
cout << "NO\n";
}
