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: Accepted)

#include <bits/stdc++.h>

#define fori(i, ini, lim) for(int i = int(ini); i < int(lim); i++)
#define ford(i, ini, lim) for(int i = int(ini); i >= int(lim); i--)

#define debug(x) cout << "> " << #x << " = " << x << endl;
#define debug_at(arr, at) cout << "> " << #arr << "[" << at << "] = " << arr[at] << endl;
#define debug_pair(p) cout << "> " << #p << " = (" << p.first << ", " << p.second << ")" << endl;

using namespace std;

int main() {
    ios_base::sync_with_stdio(false);

    int n;
    cin >> n;
    map<int, int> have;

    fori(i, 0, n) {
        int val;
        cin >> val;
        ++have[val];
    }

    int m;
    cin >> m;
    map<int, int> need;

    fori(i, 0, m) {
        int val;
        cin >> val;
        ++need[val];
    }

    bool ans = true;
    for(auto p : need) {
        if(have[p.first] < p.second) {
            ans = false;
        }
    }
    cout << (ans ? "YES" : "NO") << "\n";

    return 0;
}