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>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int a[203030],b[202020],ali[404][404],p[404];
int main()
{
	int n,m,i,ans=0,j,x,y,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++)
	{
		ali[i][i]=1;
		for(j=m;j>=1;j--)
		{
			x=a[j],y=b[j];
			if(ali[i][x]+ali[i][y]==0)
				continue;
			if(ali[i][x]+ali[i][y]==2)
				p[i]=1;
			ali[i][x]=ali[i][y]=1;
		}
	}
	for(i=1;i<=n;i++)
		for(j=i+1;j<=n;j++)
		{
			if(p[i]||p[j])
				continue;
			for(k=1;k<=n;k++)
				if(ali[i][k]&&ali[j][k])
					break;
			if(k>n)
				ans++;
		}
	printf("%d\n",ans);
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
 	return 0;
}
