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>
using namespace std;
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
//#define int long long
#define ll long long
#define ull unsigned long long
#define ld long double
#define vl vector<long long>
#define vvl vector<vector<long long>>
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define rep2(i, s, n) for (ll i = (s); i < (ll)(n); i++)
#define drep(i, x) for (ll i = ((ll)(x)-1); i >= 0; i--)
#define pll pair<long long,long long>
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define vvc vector<vector<char>>
#define vc vector<char>
#define vvb vector<vector<bool>>
#define vb vector<bool>
#define Maxe(x) *max_element(all(x))
#define Mine(x) *min_element(all(x))
#define Size(x) ((ll)(x).size())
#define umap unordered_map
const long long INF = 1LL << 60;
const long double pi = 3.1415926535897932;
#define Fi first
#define Se second
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

    template<class T> inline bool chmin(T& a, T b) {
        if (a > b) {
            a = b;
            return true;
        }
        return false;
    }
    template<class T> inline bool chmax(T& a, T b) {
        if (a < b) {
            a = b;
            return true;
        }
        return false;
    }

//lis,modpow,uf,modncr,soinsubunkai,yakusurekkyo
//mint,dijkstra,gyakugen

    signed main(){
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        map<ll,ll> D;
        ll n;
        cin>>n;
        rep(i,n){
            ll a;
            cin>>a;
            D[a]++;
        }
        ll m;
        cin>>m;
        bool res=true;
        rep(i,m){
            ll a;
            cin>>a;
            D[a]--;
            if(D[a]<0)res=false;
        }
        if(res)cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }