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 405
#define M 100005
using namespace std;
int pt[N][N],alive[N];
int f[M],t[M];
int n,m;
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
		scanf("%d%d",&f[i],&t[i]);
	for(int i=1;i<=n;i++)
		pt[i][i]=1;
	for(int i=1;i<=n;i++)
		for(int j=m;j>=1;j--){
			if(pt[i][f[j]]+pt[i][t[j]]!=0){
				if(pt[i][f[j]]+pt[i][t[j]]==2) {alive[i]=-1;break;}
				if(pt[i][f[j]]==1) pt[i][t[j]]=1;
				else pt[i][f[j]]=1;
			}
		}
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++){
			if(alive[i]==-1||alive[j]==-1) continue;
			bool fl=true;
			for(int k=1;k<=n;k++)
				if(pt[i][k]+pt[j][k]==2){
					fl=false;break;
				}
			if(fl) ans++;
		}
	cout<<ans<<endl;
	return 0;
}