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 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
int R=1,C=1,H[2000010],N[2000010];
//評価関数（いまはMIN）
int hyouka(int a,int b){
  if(C<b)return 1;
  if(C<a||b==0)return 0;
  return N[H[a]]<N[H[b]]?1:0;
}
//挿入関数
void hin(int a){
  int i=C++;
  for(N[H[0]=R]=a;hyouka(0,i/2);i/=2)H[i]=H[i/2];
  H[i]=R++;
}
//取り出す関数
int hout(){
  int rt=H[1],i,j=2,k=H[--C];
  for(i=1;hyouka(i,C);i=j)H[i]=H[j=i*2+1-hyouka(i*2,i*2+1)];
  H[j/2]=k;
  return rt;
}
int main(){
  int n,m,a,d[100010],i,r=0;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&a);
    hin(a);
  }
  for(i=0;i<n;i++)d[i]=N[hout()];
  R=C=1;
  scanf("%d",&m);
  for(i=0;i<m;i++){
    scanf("%d",&a);
    hin(a);
  }
  d[n]=-1;
  //for(i=0;i<=n;i++)printf("%d\n",d[i]);
  for(i=0;i<m;i++){
    a=N[hout()];//printf("%d\n",a);
    while(d[r]<a&&r<n)r++;
    if(d[r++]-a)goto END;
  }
  printf("YES\n");
  return 0;
 END:;
  printf("NO\n");
  return 0;
}