#include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #include <bits/stdc++.h> #define fi first #define se second #define pb push_back #define sz(a) (int)a.size() #define all(a) a.begin(), a.end() #define rall(a) a.rbegin(), a.rend() #define ff(i,a,b) for(int i=a;i<=b;i++) #define fb(i,b,a) for(int i=b;i>=a;i--) using namespace std; using namespace __gnu_pbds; typedef long long ll; typedef pair<int,int> pii; const int maxn = 21; const int inf = 1e9 + 5; template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; // os.order_of_key(k) the number of elements in the os less than k // *os.find_by_order(k) print the k-th smallest number in os(0-based) int n, m; int A[maxn]; int B[maxn]; bool dp[20][(1 << 20)]; int main() { ios::sync_with_stdio(false); cout.tie(nullptr); cin.tie(nullptr); cin >> n >> m; ff(i,0,n - 1)cin >> A[i]; ff(i,0,m - 1)cin >> B[i]; dp[0][0] = 1; ff(i,0,n - 1){ ff(mask,0,(1 << m) - 1){ // ako za prve i koristim mask, da ih dobijem sve // sad treba da vidim koje sve submaske mi daju zbir jednak A[i] int smask = mask; while(smask > 0){ int zbir = 0; ff(j,0,m - 1){ if(smask & (1 << j))zbir += B[j]; } if(zbir == A[i])dp[i + 1][mask] |= dp[i][mask ^ smask]; smask = (smask - 1) & mask; } } } ff(mask,0,(1 << m) - 1){ if(dp[n][mask] == 1)return cout << "YES", 0; } cout << "NO" << endl; return 0; } /** // probati bojenje sahovski ili slicno **/