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<cstdio>

using namespace std;

int n,m,x[100001],y[100001],ans;
bool check[401][401],c[401];

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){
		check[i][i]=true;
		for(int j=m;j;--j){
			bool b1=check[i][x[j]],b2=check[i][y[j]];
			if(b1&&b2){c[i]=true;break;}else if(b1)check[i][y[j]]=true;else if(b2)check[i][x[j]]=true;  
		}
	}
	for(int i=1;i<n;++i)if(!c[i])for(int j=i+1;j<=n;++j)if(!c[j]){
		bool flag=true;
		for(int k=1;k<=n;++k)if(check[i][k]&&check[j][k]){flag=false;break;}
		ans+=flag;
	}
	printf("%d",ans);
}