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>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;

static const double EPS = 1e-8;
static const double PI = 4.0 * atan(1.0);
static const ll INF = 1023456789;

#define REP(i,n)	for(int i=0;i<n;++i)
#define REPR(i,n)	for(int i=n-1;i>=0;--i)
#define FOR(i,s,n)	for(int i=s;i<n;++i)
#define FORR(i,s,n)	for(int i=n-1;i>=s;--i)
#define ALL(c)		(c).begin(),(c).end()
#define CLEAR(v)	memset(v,0,sizeof(v))
#define MP(a,b)		make_pair((a),(b))
#define ABS(a)		((a)>0?(a):-(a))
#define F			first
#define S			second

int n, m, d[200000], t[200000];

int main(int argc, char **argv) {
	cin >> n;
	REP(i, n) cin >> d[i];
	cin >> m;
	REP(i, m) cin >> t[i];
	sort(d, d + n);
	sort(t, t + m);
	int a = 0, b = 0;
	REP(i, m) {
		while (a < n && d[a] < t[b]) ++a;
		if (a >= n || d[a] > t[b]) break;
		++a; ++b;
	}
	cout << (b == m ? "YES" : "NO") << endl;
	return 0;
}