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 ll long long
using namespace std;
struct aaa{
	int x,y;
}a[200010];
int n,m,i,j,k,x,y,flag,ans,v[402][402],mp[402];
int main(){
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;i++)scanf("%d%d",&a[i].x,&a[i].y);
	for(i=1;i<=n;i++){
		v[i][i]=1;
		for(j=m;j;j--){
			x=a[j].x;y=a[j].y;
			if(v[i][x]&&v[i][y]){
				mp[i]=1;break;
			}
			if(v[i][x]||v[i][y])v[i][x]=v[i][y]=1;
		}
	}
	for(i=1;i<=n;i++)
		for(j=i+1;j<=n;j++){
			flag=0;
			if(mp[i]||mp[j])continue;
			for(k=1;k<=n;k++)if(v[i][k]&&v[j][k]){
				flag=1;break;
			}
			if(!flag)ans++;
		}
	printf("%d",ans);
}