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<iostream>
using namespace std;
struct tur{
	int x,y;
}q[100001];
int c[501][401],n,m,ans,a[401][401];
inline bool check(int x,int y){
	if (a[x][y]) return false;
	if (a[x][0]||a[y][0]) return false;
	for (int i=1; i<=n; i++)
		if (c[x][i]&&c[y][i]) return false;
	return true;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1; i<=m; i++)
		scanf("%d%d",&q[i].x,&q[i].y),
		a[q[i].x][q[i].y]=a[q[i].y][q[i].x]=1;
	for (int i=1; i<=n; i++){
		c[i][i]=1;
		for (int j=m; j>=1; j--){
			if (c[i][q[j].x]&&c[i][q[j].y]){a[i][0]=1; break;}
			if (c[i][q[j].x]) c[i][q[j].y]=1;
			else if (c[i][q[j].y]) c[i][q[j].x]=1;
		}
	}
	for (int i=1; i<=n; i++)
		for (int j=1+i; j<=n; j++)
			if (check(i,j)) ans++;
	printf("%d\n",ans);
	return 0;
}
