//https://oj.uz/problem/view/IZhO14_bank
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9+7;
int main()
{
//freopen("bank.in","r",stdin);
//freopen("bank.out","w",stdout);
int n, m;
cin >> m >> n;
vector< int > notes(n),salary(m);
for (int i = 0; i < m ;i++)
{
cin >> salary[i];
}
for (int i = 0; i < n ;i++)
{
cin >> notes[i];
}
vector< pair<int,int> > dp(1<<n,{-1,0});
/*
dp[00001011b] ={x,y} means select first,second and 4th notes can satisfy first x people
and also has y money left.
we add new note as the last one to cover the current people.
*/
dp[0] = {0,0} ;//empty, dealing with salary[0],no money left.
for (int s = 0; s < (1<<n); s++)
{
//cout << s << " " << dp[s].first << " " << dp[s].second << endl;
if(dp[s].first == -1) continue; //should not use invalid status.
//cout <<salary[dp[s].first] << endl;
for (int i = 0 ;i < n; i++) {
if ( (1<<i) & s ) //already has ith notes, skip
continue;
//try to finish people dp[s].first using current notes[i]
if (salary[dp[s].first] == dp[s].second + notes[i])//just enough to pay
{
dp[(1<<i) | s].first = dp[s].first+1;
dp[(1<<i) | s].second = 0;
if (dp[(1<<i) | s].first == m ) {
cout << "YES" << endl;
return 0;
}
}
else if (salary[dp[s].first] > dp[s].second + notes[i])
{
dp[(1<<i) | s].first = dp[s].first;
dp[(1<<i) | s].second = dp[s].second + notes[i];
}
//cout <<((1<<i) | s) <<" "<< dp[(1<<i) | s].first << " dp " << dp[(1<<i) | s].second << endl;
//if bigger do nothing, as we can not use it.
}
}
cout << "NO" << endl;
return 0;
}
