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 M=1e5+5;
const int N=405;
int n,m,x[M],y[M],vis[N],vi[N][N];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
		scanf("%d%d", &x[i],&y[i]);
	for(int i=1;i<=n;i++){
		vi[i][i]=m+1;
		for(int j=m;j;j--){
			if(vi[i][x[j]]&&vi[i][y[j]]){
				vis[i]=1;
				break;
			}
			if(vi[i][x[j]]||vi[i][y[j]]){
				if(!vi[i][x[j]])	vi[i][x[j]]=j;
				if(!vi[i][y[j]])	vi[i][y[j]]=j;
			}
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++){
		if(vis[i])	continue;
		for(int j=i+1;j<=n;j++){
			if(vis[j])	continue;

			int f=1;
			for(int k=1;k<=n;k++)
				if(vi[i][k]&&vi[j][k]&&vi[i][k]!=vi[j][k]){
					//cout<<i<<' '<<j<<' '<<k<<' '<<vi[i][k]<<' '<<vi[j][k]<<endl;
					f=0;
				}
			ans+=f;
		}
	}
	printf("%d\n",ans);
	return 0;
}