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>
using namespace std;
int n,m,x[100010],y[100010],i,j,k,ans[410],s[410][410];
int main()
{
	scanf("%d%d",&n,&m);
	for (i=1;i<=m;i++) scanf("%d%d",&x[i],&y[i]);
	for (i=1;i<=n;i++)
	{
		ans[i]=1;
		s[i][i]=1;
		for (j=m;j>=1;j--)//need to survive before the jth round?
		{
			if (s[i][x[j]]==0&&s[i][y[j]]==0) continue;
			if (s[i][x[j]]==1&&s[i][y[j]]==1) 
			{
				ans[i]=0;
				break;
			}
			s[i][x[j]]=1;s[i][y[j]]=1;
		}
	}
	int now=0;
	for (i=1;i<=n;i++)
	  for (j=i+1;j<=n;j++)
	  {
	  	if (ans[i]==0||ans[j]==0) continue;
	  	now++;
	  	for (k=1;k<=n;k++)
	  	  if (s[i][k]+s[j][k]>1)
	  	  {
	  	  	now--;
	  	  	break;
		  }
	  }
	printf("%d\n",now);
 } 