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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define FOR(i, a, b) for (int i = a; i <= (b); i++)
#define ROF(i, a, b) for (int i = a; i >= (b); i--)

using pii = pair<int, int>; using vpii = vector<pii>;
using vi = vector<int>; using vvi = vector<vi>;
using ll = long long;
using pll = pair<ll, ll>; using vpll = vector<pll>;
using vll = vector<ll>; using vvll = vector<vll>;

int main() {
    int n, m; cin >> n >> m;
    map <int, int> cnt;
    FOR(i, 1, n) {
        int tt; cin >> tt;
        cnt[tt]++;
    }
    FOR(i, 1, m) {
        int tt; cin >> tt;
        cnt[tt]--;
        if (cnt[tt] < 0) return cout << "NO\n", 0;
    }
    cout << "YES\n";
}