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<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

const int maxn=410;
const int maxm=100010;
int n, m;
int x[maxm], y[maxm];
int a[maxn][maxn];

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<=m;i++){
		for(int j=1;j<=n;j++) if(a[ x[i] ][j]) a[ y[i] ][j]=a[j][ y[i] ]=1;
		for(int j=1;j<=n;j++) if(a[ y[i] ][j]) a[ x[i] ][j]=a[j][ x[i] ]=1;
		if(a[ x[i] ][ y[i] ]){
			for(int j=1;j<=n;j++) a[ x[i] ][j]=a[j][ x[i] ]=1, a[ y[i] ][j]=a[j][ y[i] ]=1;
		}
		a[ x[i] ][ y[i] ]=a[ y[i] ][ x[i] ]=1;
	}
	int ans=n*(n-1)>>1;
	for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++) ans-=a[i][j];
	printf("%d\n", ans);
	return 0;
}