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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; ++i)
#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define ALL(vec) vec.begin(), vec.end()
#define rALL(vec) vec.rbegin(), vec.rend()
using namespace std;
using ll = long long;

int main() {
    int N;
    cin >> N;
    vector<int> D(N);
    rep(i, N) cin >> D[i];
    int M;
    cin >> M;
    vector<int> T(M);
    rep(i, M) cin >> T[i];
    sort(ALL(D));
    sort(ALL(T));
    int idx = 0;
    rep(i, M) {
        bool flag = true;
        FOR(j, idx, N) if (T[i] == D[j]) {flag = false; idx = j + 1; break;}
        if (flag) {cout << "NO" << endl; return 0;}
    }
    cout << "YES" << endl;
    return 0;
}