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<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<bitset>
using namespace std;
int pd[410],u[110000],v[110000],n,m;
bitset<410>A[410],x[110000],ans,y[110000];
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++) scanf("%d%d",&u[i],&v[i]);
	for (int i=1;i<=n;i++) A[i][i]=1;
	for (int i=m;i;i--){
		x[i]=A[u[i]];
		y[i]=A[v[i]];
		A[u[i]]|=A[v[i]];
		A[v[i]]=A[u[i]];
	}
	int num=0;
	for (int i=1;i<=m;i++)
		for (int j=1;j<=n;j++)
			if (x[i][j]&&y[i][j]) pd[j]=1;
	for (int i=1;i<=n;i++){
		ans=0; if (pd[i]) continue;
		for (int j=1;j<=m;j++){
			if (x[j][i]) ans|=y[j];
			if (y[j][i]) ans|=x[j];
		}
		for (int j=i+1;j<=n;j++) if (ans[j]==0&&pd[j]==0) num++;
	//	for (int j=1;j<=n;j++) cout<<ans[j]; cout<<endl;
	}
	cout<<num<<endl;
	return 0;
}