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>
#define pb push_back

using namespace std;

const int N=100005,M=405;

bitset<M>v[M];
bool dead[M];
int n,m,ans;

int main(){
	scanf("%d%d",&n,&m);
	for(int i=1,x,y;i<=m;i++){
		scanf("%d%d",&x,&y);
		if(dead[x]||dead[y]||(v[x]&v[y]).any()){dead[x]=dead[y]=true;continue;}
		v[x]=v[y]=v[x]|v[y];
		if(v[x][x])dead[x]=true;if(v[y][y])dead[y]=true;
		v[x][y]=1,v[y][x]=1;
	}
	for(int x=1;x<=n;x++)if(!dead[x])for(int y=x+1;y<=n;y++)if(!dead[y])
		if(!v[x][y]&&!v[y][x]&&!(v[x]&v[y]).any())ans++;
	printf("%d\n",ans);
	return 0;
}