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<bits/stdc++.h>
#define L long long
using namespace std;
int n,m,x[410][410],a[100010],b[100010],p;
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k;
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;i++)
	  scanf("%d%d",&a[i],&b[i]);
	for(i=1;i<=n;i++)
	  {
	   x[i][i]=1;
	   for(j=m;j>0;j--)
	     if(x[i][a[j]])
	       if(x[i][b[j]])
	         x[i][0]=1;
	       else
	         x[i][b[j]]=1;
	     else
	       if(x[i][b[j]])
	         x[i][a[j]]=1;
	  }
	for(i=1;i<=n;i++)
	  for(j=i+1;j<=n;j++)
	    if(!x[i][0] && !x[j][0])
	      {
		   for(k=1;k<=n;k++)
		     if(x[i][k] && x[j][k])
		       break;
		   if(k>n)
		     p++;
		  }
	printf("%d\n",p);
	return 0;
}
