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;
#define M 100010
#define N 410
int n,m,a[M],b[M],f[N][N],c[N];
int main(){
	cin>>n>>m;
	for(int i=1;i<=m;i++)
		cin>>a[i]>>b[i];
	for(int i=1;i<=n;i++){
		f[i][i]=1;
		for(int j=m;j;j--){
			int x=f[i][a[j]],y=f[i][b[j]];
			if(x&&y){
				c[i]=1;
				break;
			}
			else if(x)f[i][b[j]]=1;
			else if(y)f[i][a[j]]=1;
		}
	}
	int r=0;
	for(int i=1;i<n;i++){
		if(c[i])continue;
		for(int j=i+1;j<=n;j++){
			if(c[j])continue;
			int ok=1;
			for(int k=1;k<=n;k++)
				if(f[i][k]&&f[j][k])ok=0;
			r+=ok;
		}
	}
	cout<<r;
}