#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
using namespace std;
int N, M;
int a[21];
int b[21];
vector<int> masks_by_sum[1001]; // Only need up to 1000 as per constraints
int memo[1 << 20]; // Bitmask to store which persons (p_idx) failed for a banknote mask
bool solve(int p_idx, int b_mask) {
// If we have satisfied all N people, we are done
if (p_idx == N) return true;
// If this specific combination of banknotes has already failed for this person, skip
if (memo[b_mask] & (1 << p_idx)) return false;
int target = a[p_idx];
// Try all precalculated masks that sum up to the current person's salary
for (int m : masks_by_sum[target]) {
// Check if these banknotes are still available (not in b_mask)
if (!(b_mask & m)) {
if (solve(p_idx + 1, b_mask | m)) return true;
}
}
// Mark this state as failed
memo[b_mask] |= (1 << p_idx);
return false;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
if (!(cin >> N >> M)) return 0;
long long sum_a = 0;
for (int i = 0; i < N; ++i) {
cin >> a[i];
sum_a += a[i];
}
long long sum_b = 0;
for (int i = 0; i < M; ++i) {
cin >> b[i];
sum_b += b[i];
}
// Basic feasibility check
if (sum_a > sum_b) {
cout << "NO" << endl;
return 0;
}
// Sorting salaries descending helps prune the search tree significantly
sort(a, a + N, greater<int>());
// Precalculate all banknote subset sums that match any a_i
bool needed[1001] = {false};
for(int i=0; i<N; ++i) needed[a[i]] = true;
for (int i = 0; i < (1 << M); ++i) {
int current_sum = 0;
for (int j = 0; j < M; ++j) {
if (i & (1 << j)) {
current_sum += b[j];
if (current_sum > 1000) break;
}
}
if (current_sum <= 1000 && needed[current_sum]) {
masks_by_sum[current_sum].push_back(i);
}
}
if (solve(0, 0)) cout << "YES" << endl;
else cout << "NO" << endl;
return 0;
}
