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>
bool a[405][405],u[405];
int x[100005],y[100005],n,m;
bool check(int i,int j){
	if (!u[i] || !u[j]) return 0;
	for (int k=1;k<=n;k++)
		if (a[i][k] && a[j][k]) return 0;
	return 1;
}
int main(){
	scanf("%d%d\n",&n,&m);
	for (int i=1;i<=m;i++) scanf("%d%d\n",&x[i],&y[i]);
	for (int i=1;i<=n;i++){
		a[i][i]=u[i]=1;
		for (int j=m;j && u[i];j--)
		if (a[i][x[j]] && a[i][y[j]]) u[i]=0;
		else if (a[i][x[j]] || a[i][y[j]]) a[i][y[j]]=a[i][x[j]]=1;
	}
	int ans=0;
	for (int i=1;i<n;i++)
		for (int j=i+1;j<=n;j++) ans+=check(i,j);
	printf("%d\n",ans);
}