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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#include <functional>
#include <queue>
//#include <stack>
#include <climits>
#include <map>
#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(i,n1,n2) for(int i=n1;i<n2;i++)
#define put(a) cout<<a<<endl;
using namespace std;
typedef long long ll;
//typedef unsigned long long ull;
typedef pair<int, int> P;

int n,m;
map<int,int> mp1;
map<int,int> mp2;

int main() {
    cin >> n;
    vector<int> d(n);
    REP(i,n){
        cin >> d[i];
    }
    cin >> m;
    vector<int> t(m);
    REP(i,m){
        cin >> t[i];
    }
    //map<int,int> m;
    REP(i,n){
        auto itr = mp1.find(d[i]);
        if(itr==mp1.end()){
            mp1[d[i]]=1;
        }else{
            mp1[d[i]]++;
        }
    }
    REP(i,m){
        auto itr = mp2.find(t[i]);
        if(itr==mp2.end()){
            mp2[t[i]]=1;
        }else{
            mp2[t[i]]++;
        }
    }
    REP(i,m){
        auto itr = mp1.find(t[i]);
        if(itr==mp1.end()){
            put("NO");
            return 0;
        }else{
            if(mp1[t[i]]==0){
                put("NO");
                return 0;
            }else{
                mp1[t[i]]--;
            }
        }
    }
    put("YES");
    return 0;
}