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<stdio.h>
#include<string.h>
using namespace std;
long long g[420][12];
int x[120000],y[120000],p[420];
int n,m,ans;
bool find(int o,int x){
	int k=x/50;x%=50;
	return (g[o][k]&(1LL<<x))>0;
}
int add(int o,int x){
	int k=x/50;x%=50;
	g[o][k]|=1LL<<x;
}
int solve(int o){
	add(o,o);
	for(int i=m;i>=1;i--){
		int fx=find(o,x[i]),fy=find(o,y[i]);
		if(fx+fy==0)continue;
		if(fx+fy==2){
			p[o]=1;
			return 0;
		}
		if(fx==0)add(o,x[i]);
		if(fy==0)add(o,y[i]);
	}
}
int cal(int x,int y){
	for(int i=0;i<=8;i++)if(g[x][i]&g[y][i])return 0;
	return 1;
}
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++)solve(i);
//	for(int i=1;i<=n;i++)printf("%d\n",g[i][0]);
	for(int i=1;i<=n;i++){
		if(p[i])continue;
		for(int j=i+1;j<=n;j++){
			if(p[j])continue;
			ans+=cal(i,j);
	//		if(cal(i,j)==1)
	//		printf("%d %d\n",i,j);
		}
	}
	printf("%d\n",ans);
}