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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <stdbool.h>
typedef long long LL;
typedef long double Lf;
#define Rep(i,a,n)for(int i=(int)(a);i<(int)(n);i++)
//#define Rep(i,a,n)for(LL i=(LL)(a);i<(LL)(n);i++)
#define rep(i,n)Rep(i,0,n)
#define Repp(i,l,r,k)for(int i=(int)(l);i<(int)(r);i+=(int)(k))
//#define Repp(i,l,r,k)for(LL i=(LL)(l);i<(LL)(r);i+=(LL)(k))
#define Ma(x,y) ((x)>(y)?(x):(y))
#define mi(x,y) ((x)<(y)?(x):(y))
#define MOD 1000000007//998244353,1000000007,1000000009 
int upi(const void* a, const void* b) { return*(int*)a<*(int*)b ? -1 : *(int*)a> * (int*)b ? 1 : 0; }
int downi(const void* a, const void* b) { return*(int*)a<*(int*)b ? 1 : *(int*)a> * (int*)b ? -1 : 0; }
void sortupi(int* a, int n) { qsort(a, n, sizeof(int), upi); } //1,2,3,4,...
void sortdowni(int* a, int n) { qsort(a, n, sizeof(int), downi); }
int upLL(const void* a, const void* b) { return*(LL*)a<*(LL*)b ? -1 : *(LL*)a> * (LL*)b ? 1 : 0; }
int downLL(const void* a, const void* b) { return*(LL*)a<*(LL*)b ? 1 : *(LL*)a> * (LL*)b ? -1 : 0; }
void sortupLL(LL* a, int n) { qsort(a, n, sizeof(LL), upLL); }
void sortdownLL(LL* a, int n) { qsort(a, n, sizeof(LL), downLL); }
int cmp_char(const void* a, const void* b) { return *(char*)a - *(char*)b; } // a,b,c,d
int cmp_str(const void* a, const void* b) { return strcmp(*(const char**)a, *(const char**)b); } // aaa,aab
LL gcd(LL p, LL q) { while (q) { LL t = p % q; p = q; q = t; }return p; }
LL modpow(LL a,LL n,int m){LL x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
#define invmodpow(b,p)modpow(b,p-2,p)// a/b≡a*b^{-1}(mod p)...return b^{-1}(mod p)
typedef struct fac { LL n, exp; }F;
LL fact(LL n,F*a){LL x,i,j,c=0;x=n;for(i=2;i*i<=x;i++){for(j=1;x%i==0;j++)x/=i;if(j>1){a[c].n=i;a[c++].exp=j-1;}}if(x-1){a[c].n=x;a[c++].exp=1;}return c;}
//snippet
//overflow...char 1e6

#define N 200009

int main() {
	int n, d[N], m, t[N];
	scanf("%d", &n);
	rep(i, n)scanf("%d", &d[i]);
	scanf("%d", &m);
	rep(i, m)scanf("%d", &t[i]);
	sortupi(d, n);

	rep(i, m) {
		//にぶたん
		int p = 0, r = n - 1, q = n / 2, f = 0;
		while (r - p > 2) {
			if (t[i] <= d[q])r = q;
			if (t[i] >= d[q])p = q;
			q = (p + r) / 2;
		}
		if (t[i] == d[p]) {
			f = 1;
			d[p] = 0;
		}
		if (t[i] == d[q]) {
			f = 1;
			d[q] = 0;
		}
		if (t[i] == d[r]) {
			f = 1;
			d[r] = 0;
		}
		sortupi(d, n);
		if (!f) {
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");


	return 0;
}