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<cstring>
#define M 100005
#define N 404
int a[M],b[M],m,n,x;bool w[N][N];
inline bool solve(int u)
{
	w[u][u]=true;
	for(int i=m;i;i--)
	if(w[u][a[i]]&&w[u][b[i]])return false;
	else if(w[u][a[i]])
	{
		if(w[u][b[i]])return false;
		w[u][b[i]]=true;
	}
	else if(w[u][b[i]])
	{
		if(w[u][a[i]])return false;
		w[u][a[i]]=true;
	}
	return true;
}
inline bool check(int u,int v)
{
	for(int i=1;i<=n;i++)if(w[u][i]&w[v][i])return false;
	return true;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)scanf("%d%d",a+i,b+i);
	for(int i=1;i<=n;i++)w[i][i]=solve(i);
	for(int i=1;i<n;i++)if(w[i][i])for(int j=i+1;j<=n;j++)if(w[j][j])x+=check(i,j);
	return 0*printf("%d\n",x);
}