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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

type
  arr=array[1..200000]of longint;
var
  n,i,m,j,x:longint;
  d,t:arr;

  procedure sort(l,r:longint;var a:arr);
  var
    i,j,x,y:longint;
  begin
    i:=l; j:=r;
    x:=a[(l+r) div 2];
    while i<=j do
      begin
        while a[i]<x do inc(i);
        while x<a[j] do dec(j);
        if i<=j then begin y:=a[i]; a[i]:=a[j]; a[j]:=y; inc(i); dec(j); end;
      end;
    if l<j then sort(l,j,a);
    if i<r then sort(i,r,a);
  end;

begin
  readln(n);
  for i:=1 to n do read(d[i]);
  sort(1,n,d);
  readln(m);
  for i:=1 to m do read(t[i]);
  sort(1,m,t);
  i:=1; j:=1;
  while (i<=n)and(j<=m) do
    begin
      x:=d[i];
      while (d[i]=t[j])and(d[i]=x)and(i<=n)and(j<=m) do begin inc(i); inc(j); end;
      while (d[i]<t[j])and(i<=n) do inc(i);
      if j=m+1 then begin writeln('YES'); halt; end;
      if d[i]<>t[j] then begin writeln('NO'); halt; end;
    end;
  writeln('NO');
end.