There are N turkeys. We number them from 1 through N.

M men will visit here one by one. The i-th man to visit will take the following action:

* If both turkeys x_i and y_i are alive: selects one of them with equal probability, then eats it.
* If either turkey x_i or y_i is alive (but not both): eats the alive one.
* If neither turkey x_i nor y_i is alive: does nothing.



Find the number of pairs (i,\ j) (1 ≤ i < j ≤ N) such that the following condition is held:

* The probability of both turkeys i and j being alive after all the men took actions, is greater than 0.

Constraints

* 2 ≤ N ≤ 400
* 1 ≤ M ≤ 10^5
* 1 ≤ x_i < y_i ≤ N

Input

Input is given from Standard Input in the following format:


N M
x_1 y_1
x_2 y_2
:
x_M y_M


Output

Print the number of pairs (i,\ j) (1 ≤ i < j ≤ N) such that the condition is held.

Examples

Input

3 1
1 2


Output

2


Input

4 3
1 2
3 4
2 3


Output

1


Input

3 2
1 2
1 2


Output

0


Input

10 10
8 9
2 8
4 6
4 9
7 8
2 8
1 8
3 4
3 4
2 7


Output

5
#include <cstdio>

int mex[500][500];

int main(){
  int N,M;
  scanf("%d %d",&N,&M);
  for(int i=0;i<M;i++){
    int X,Y;
    scanf("%d %d",&X,&Y);
    X--,Y--;
    if(mex[X][X]||mex[Y][Y]||mex[X][Y]){
      mex[X][X]=1;
      mex[Y][Y]=1;
    }else{
      for(int j=0;j<N;j++){
	if(mex[X][j]||mex[Y][j]){
	  mex[X][j]=mex[j][X]=mex[Y][j]=mex[j][Y]=1;
	}
      }
      mex[X][Y]=mex[Y][X]=1;
    }
  }
  for(int i=0;i<N;i++){
    if(mex[i][i]){
      for(int j=0;j<N;j++){
	mex[i][j]=mex[j][i]=1;
      }
    }
  }
  /*
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++){
      printf("%d",mex[i][j]);
    }
    printf("\n");
  }
  */
  int cnt=0;
  for(int i=0;i<N;i++){
    for(int j=0;j<i;j++){
      if(mex[i][j]==0) cnt++;
    }
  }
  printf("%d\n",cnt);
  return 0;
}
