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 N 409
#define M 100009
using namespace std;

int n,m,a[M],b[M]; bool bo[N][N],ban[N];
int main(){
	scanf("%d%d",&n,&m);
	int i,j,k,ans=0;
	for (i=1; i<=m; i++) scanf("%d%d",&a[i],&b[i]);
	for (i=1; i<=n; i++){
		bo[i][i]=1;
		for (j=m; j; j--) if (bo[i][a[j]] && bo[i][b[j]]){
			ban[i]=1; break;
		} else if (bo[i][a[j]]) bo[i][b[j]]=1;
		else if (bo[i][b[j]]) bo[i][a[j]]=1;
	}
	for (i=1; i<n; i++) if (!ban[i])
		for (j=i+1; j<=n; j++) if (!ban[j]){
			for (k=1; k<=n; k++) if (bo[i][k] && bo[j][k]) break;
			if (k>n) ans++;
		}
	printf("%d\n",ans);
	return 0;
}
