//https://oj.uz/problem/view/IZhO14_bank
#include <bits/stdc++.h>
using namespace std;
vector<int> salaries;
vector<int> notes;
vector<vector<int>> sums;
int main() {
cin.tie(0) -> sync_with_stdio(0);
int N, M; cin >> N >> M;
//use bitmask dp on bitmask dp :skull:
//precalculate all the different bitmasks required for each individual one
//then when we run a xor we can see if there is any overlap
salaries.resize(N);
notes.resize(M);
sums.resize(1001);
for (int i = 0; i < N; i++) {
cin >> salaries[i];
}
for (int i = 0; i < M; i++) {
cin >> notes[i];
}
//find the bitmasks
for (int mask = 0; mask < 1 << M; mask++) {
int sum = 0;
for (int b = 0; b < M; b++) {
if (!(mask & 1 << b)) { continue; }
sum += notes[b];
}
if (sum > 1000) { continue; }
sums[sum].push_back(mask);
}
//now the hard part: form all possible stuff
//i am trll
//im betting its a better idea to use sets here?
vector<unordered_set<int>> possible(1 << N); //bitmasks which are possible, with bitmasks of notes
possible[0].insert(0);
for (int mask = 1; mask < 1 << N; mask++) {
for (int b = 0; b < N; b++) {
if (!(mask & 1 << b)) { continue; }
int from = mask ^ 1 << b;
for (int new_m : sums[salaries[b]]) {
for (int old_m : possible[from]) { //ways to get to old
if (new_m & old_m) { continue; } //if overlap goodbye
possible[mask].insert(new_m ^ old_m);
}
}
}
}
bool valid = possible[(1 << N) - 1].size() > 0;
cout << (valid ? "YES" : "NO") << "\n";
}
