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 maxn=5e2+10,maxm=1e5+10;
int u[maxm],v[maxm],f[maxn][maxn],p[maxn],ans,n,m;
int main(){
	cin>>n>>m;
	for(register int i=1;i<=m;++i)scanf("%d%d",&u[i],&v[i]);
	for(register int i=1;i<=n;++i){
		f[i][i]=1;
		for(register int j=m;j;--j){
			if(f[i][u[j]]&&f[i][v[j]])break;
			if(f[i][u[j]]||f[i][v[j]])f[i][u[j]]=f[i][v[j]]=1;
			if(j==1)p[i]=1;
		}
	}for(register int i=1;i<=n;++i)if(p[i])for(register int j=i+1;j<=n;++j)if(p[j]){
		int tmp=1;
		for(register int k=1;k<=n;++k)if(f[i][k]&&f[j][k]){tmp=0;break;}
		ans+=tmp;
	}cout<<ans<<endl;
	return 0;
}