#include <bits/stdc++.h>
using namespace std;
int main() {
// Optimize standard I/O operations for performance
ios::sync_with_stdio(0); cin.tie(0);
int n, m;
if (!(cin >> n >> m)) return 0;
vector<int> people(n), money(m);
for (int i = 0; i < n; i++) cin >> people[i];
for (int i = 0; i < m; i++) cin >> money[i];
// dp[mask] stores a pair:
// .first = maximum number of people fully paid off
// .second = money accumulated toward the NEXT person's salary
// We initialize with {-1, -1} to flag unreachable states.
vector<pair<int, int>> dp(1 << m, {-1, -1});
// Base case: 0 banknotes used, 0 people paid, 0 money accumulated
dp[0] = {0, 0};
for (int mask = 0; mask < (1 << m); mask++) {
// Skip states that cannot be reached
if (dp[mask].first == -1) continue;
int p_paid = dp[mask].first;
int curr_sum = dp[mask].second;
// If we have successfully paid all N people, we can stop immediately
if (p_paid == n) {
cout << "YES\n";
return 0;
}
// Try adding each unused banknote to the current mask
for (int i = 0; i < m; i++) {
// Check if the i-th bit is OFF (banknote 'i' is not used yet)
if (!(mask & (1 << i))) {
int nmask = mask | (1 << i); // Create the new state mask
int nsum = curr_sum + money[i]; // Add the banknote's value
// Case 1: The accumulated money is still less than the current person's salary
if (nsum < people[p_paid]) {
dp[nmask] = max(dp[nmask], {p_paid, nsum});
}
// Case 2: The accumulated money EXACTLY matches the current person's salary
else if (nsum == people[p_paid]) {
// Increment people paid, reset the accumulated money to 0
dp[nmask] = max(dp[nmask], {p_paid + 1, 0});
}
// If nsum > people[p_paid], the banknote is too big. We do nothing (invalid transition).
}
}
}
// Final check: Iterated through all combinations but couldn't pay everyone
cout << "NO\n";
return 0;
}
