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 _=1e5+5;
int a[_],b[_],c[_],n,m;
bitset<405>A[405];
int main(){
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=1;i<=m;++i)cin>>a[i]>>b[i];
	for(int i=1;i<=n;++i){
		A[i][i]=1;
		for(int j=m;j;--j){
			if(A[i][a[j]]&&A[i][b[j]]){c[i]=1;break;}
			if(A[i][a[j]])A[i][b[j]]=1;
			if(A[i][b[j]])A[i][a[j]]=1;
		}
	}
	int ans=0;
	for(int i=1;i<=n;++i)
		for(int j=i+1;j<=n;++j)
			ans+=(!(A[i]&A[j]).count()&&!c[i]&&!c[j]);
	cout<<ans<<endl;
}
