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>
using namespace std;
const int N=402,M=100002;
int n,m,i,j,ans,x[M],y[M],k;
bool vis[N],mp[N][N],fl;
int main(){
	scanf("%d%d",&n,&m);
	for (i=1;i<=m;i++) scanf("%d%d",&x[i],&y[i]);
	for (i=1;i<=n;i++){
		mp[i][i]=1;
		for (j=m;j;j--){
			bool &xx=mp[i][x[j]],&yy=mp[i][y[j]];
			if (xx && yy){
				vis[i]=1;
				break;
			}else if (xx) yy=1;
			else if (yy) xx=1;
		}
	}
	for (i=1;i<=n;i++) if (!vis[i])
		for (j=i+1;j<=n;j++) if (!vis[j] && !mp[i][j]){
			fl=1;
			for (k=1;k<=n;k++)
				if (mp[i][k] && mp[j][k]) fl=0;
			ans+=fl;
		}
	printf("%d",ans);
}
