#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
#define MOD 1000000007
#define nl "\n"
#define pb push_back
int n, m;
bool visited[(1 << 20)]{};
bool dp[(1 << 20)];
int salariesFilled[(1 << 20)];
int lastSalaryMoney[(1 << 20)];
int salaries[20];
int banknotes[20];
bool canPaySalaries(int mask) {
if (mask == (1 << m) - 1) { // all banknotes have been used
return (salariesFilled[mask] == n - 1);
} else if (visited[mask]) {
return dp[mask];
}
visited[mask] = true;
int filled = salariesFilled[mask];
int money = lastSalaryMoney[mask];
for (int i = 0; i < m; i++) {
if (!(mask & (1 << i))) { // banknote not in mask, we can try putting it in
if (money + banknotes[i] == salaries[filled] && filled == n - 1) {
dp[mask] = true;
return true;
} else if (money + banknotes[i] <= salaries[filled]) {
if (canPaySalaries(mask + (1 << i))) {
dp[mask] = true;
return true;
}
}
}
}
return false;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n >> m;
for (int i = 0; i < n; i++) {
cin >> salaries[i];
}
for (int i = 0; i < m; i++) {
cin >> banknotes[i];
}
for (int mask = 0; mask < (1 << m); mask++) {
// we need to precalculate the number of salaries that each mask will fill, and the remaining money that will be put towards filling the next salary
// it's assumed that each salary is able to be perfectly filled, as this is the only case in which we're using the precalculated values
int sum = 0;
salariesFilled[mask] = 0;
for (int i = 0; i < m; i++) {
if (mask & (1 << i)) {
sum += banknotes[i];
}
}
for (int i = 0; i < n; i++) {
if (salaries[i] > sum) {
lastSalaryMoney[mask] = sum;
break;
} else {
sum -= salaries[i];
salariesFilled[mask]++;
}
}
}
cout << (canPaySalaries(0) ? "YES" : "NO") << nl;
}
