#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <numeric>
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define FOR(a, b) for(ll i=a; i < (ll) b; i++)
#define endl '\n'
#define MOD 1000000007
#define TESTCASE ll t; cin >> t; for(ll T=0; T < t; T++)
using namespace std;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<int> vi;
typedef pair<int, int> pi;
typedef vector<pi> vpi;
const ll INF = (ll) 1e18;
const ll INIT = 7;
const ll MAX_VAL = (ll) 34;
const ll MAX_SZ = (ll) 2e5;
const ll MULT = (ll) 1e11;
const double eps = 1e-5;
vi rd = {0, 0 , 1, -1}, cd = {1, -1, 0 , 0};
int main() {
fast
// freopen("bank.in", "r", stdin);
// freopen("bank.out", "w", stdout);
// TESTCASE {
// }
ll n,m;
cin >> n >> m;
vll salaries(n + 1), notes(m + 1), pref_sal(m + 1, 0);
FOR(1, n + 1) {
cin >> salaries[i];
pref_sal[i] = salaries[i];
pref_sal[i] += pref_sal[i - 1];
}
FOR(1, m + 1) {
cin >> notes[i];
}
ll subsets = (1 << m) - 1;
vpll dp(subsets + 1, make_pair(0, 0));
//.first tracks the number of salaries that are paid, .second tracks the current sum of banknotes
ll largest=0;
for(ll cur_subset = 1; cur_subset <= subsets; cur_subset++) {
for(ll bit=1; bit <= m; bit++) {
if(cur_subset & (1 << (bit - 1))) {
ll without = cur_subset ^ (1 << (bit - 1));
dp[cur_subset].second = dp[without].second + notes[bit];
dp[cur_subset].first = max(dp[cur_subset].first , dp[without].first + (dp[cur_subset].second - pref_sal[dp[without].first] == salaries[dp[without].first + 1]));
largest = max(largest, dp[cur_subset].first);
// cout << cur_subset << " " << bit << " " << dp[cur_subset].first << endl;
if(largest == n) {
cout << "YES";
return 0;
}
}
}
}
cout << "NO";
return 0;
}
/*
let dp[i] be the maximum number of salaries from the start which have been fulfilled using the subset i
transitions :
dp[a] = max(dp[a without element j] + (sum of all b[i] where i is in a - b[dp[a without element j]] == 0))
011111
*/
