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: Time Limit Exceeded)

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int Maxn = 200010;
int n,k,x,d[Maxn],m[Maxn],t[Maxn];
bool f;
int main()
{
    scanf("%d",&n);

    memset(d,0,sizeof(d));
    memset(m,0,sizeof(m));

    f = true;

    for (int i = 0; i < n; i++)
    {
        scanf("%d",&x);
        d[i]=x;
    }

    scanf("%d",&k);

    for (int i = 0; i < k; i++)
    {
        scanf("%d",&x);
        m[i] = x;
    }

    //sort(d,d+n);
    //sort(m,m+k);

    // for (int i = 0; i < k; i++)
    //     if(m[i] != d[i])
    //     {
    //         f = false;
    //         break;
    //     }

    for (int i = 0; i < k; i++)
        for (int j = 0; j < n; j++)
            if(m[i] == d[j])
            {
                m[i] = d[j] = 0;
                continue;
            }
    for (int i = 0; i < k; i ++)
            if(m[i] != 0)
            {
                f = false;
                break;
            }
    if(f)
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}