Problem:
Score : 200 points

Problem Statement

Rng is preparing a problem set for a qualification round of CODEFESTIVAL.

He has N candidates of problems. The difficulty of the i-th candidate is D_i.

There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems.

Determine whether Rng can complete the problem set without creating new candidates of problems.

Constraints

1 \leq N \leq 200,000

1 \leq D_i \leq 10^9

1 \leq M \leq 200,000

1 \leq T_i \leq 10^9

All numbers in the input are integers.

Partial Score

100 points will be awarded for passing the test set satisfying N \leq 100 and M \leq 100.

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N
M
T_1 T_2 ... T_M

Output

Print YES if Rng can complete the problem set without creating new candidates of problems; print NO if he cannot.

Sample Input 1

5
3 1 4 1 5
3
5 4 3

Sample Output 1

YES

Sample Input 2

7
100 200 500 700 1200 1600 2000
6
100 200 500 700 1600 1600

Sample Output 2

NO

Not enough 1600s.

Sample Input 3

1
800
5
100 100 100 100 100

Sample Output 3

NO

Sample Input 4

15
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
9
5 4 3 2 1 2 3 4 5

Sample Output 4

YES

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int (i) = 0; (i) < (n); ++(i))
#define rrep(i,n) for(int (i) = (n) - 1; (i) >= 0; --(i)) //Reversi rep
#define nfor(i,a,b) for(int (i) = (a); (i) < (b); ++(i)) //Natural for
#define rfor(i,a,b) for(int (i) = (b) - 1; (i) >= (a); --(i)) //Reversi for
#define ll long long
#define mod (1000000007)
#define YES cout << "YES" << endl;
#define Yes cout << "Yes" << endl;
#define yes cout << "yes" << endl;
#define NO cout << "NO" << endl;
#define No cout << "No" << endl;
#define no cout << "no" << endl;

int asc(const void *a, const void *b) {
    return *(int *)a - *(int *)b;
}

int desc(const void *a, const void *b) {
    return *(int *)b - *(int *)a;
}

int main()
{
    int n,m;
    ll b;
    cin >> n;
    vector<ll> d(mod,0),a(n);
    rep(i, n){
        cin >> a[i];
        ++d[a[i]];
    }
    cin >> m;
    rep(i, m){
        cin >> b;
        if (d[b] == 0) {
            NO;
            return 0;
        }else{
            --d[b];
        }
    }
    YES;
    return 0;
}