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: Wrong Answer)

var
    n,m,i:longint;
    d,t:array[-1..200000]of longint;
    td,tt:longint;

procedure sort( l,r:longint);
var
  x,e,j,i:longint;
begin
  i:=l;j:=r;x:=(l+r)div 2;
  repeat
    while (d[i]<d[x]) do inc(i);
    while(d[j]>d[x]) do dec(j);
    if(i<=j)then
    begin
      e:=d[i];d[i]:=d[j];d[j]:=e;inc(i);dec(j);end;
  until(i>j);
  if i<r then sort(i,r);
  if l<j then sort(l,j);
end;

procedure qsort( l,r:longint);
var
  x,e,j,i:longint;
begin
  i:=l;j:=r;x:=(l+r)div 2;
  repeat
    while (t[i]<t[x]) do inc(i);
    while(t[j]>t[x]) do dec(j);
    if(i<=j)then
    begin
      e:=t[i];t[i]:=t[j];t[j]:=e;inc(i);dec(j);end;
  until(i>j);
  if i<r then qsort(i,r);
  if l<j then qsort(l,j);
end;

begin
  readln(n);
  for i:=1 to n do
    read(d[i]);
  readln;
  readln(m);
  for i:=1 to m do
    read(t[i]);
  readln;
  sort(1,n);
  qsort(1,m);
  td:=n;tt:=m;
  repeat
    if t[tt]=d[td] then begin dec(tt);dec(td);end
    else if t[tt]<d[td] then dec(td)
    else if(td<tt)then begin writeln('NO');exit;end;
  until(tt=0);
  writeln('YES');
end.