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>
using namespace std;

int n,m,ex[100010],ey[100010],ans;
bool bo[410][410],boo[410],b[410];

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1; i<=m; i++) scanf("%d%d",&ex[i],&ey[i]);
	for (int i=1; i<=n; i++)
	{
		bo[i][i]=1;
		for (int j=m; j; j--)
			if (bo[i][ex[j]]||bo[i][ey[j]]) bo[i][ex[j]]=bo[i][ey[j]]=1;
		memset(boo,0,sizeof(boo)),boo[i]=1,b[i]=1;
		for (int j=m; j; j--)
		{
			if (boo[ex[j]]&&boo[ey[j]]) {b[i]=0; break;}
			if (boo[ex[j]]||boo[ey[j]]) boo[ex[j]]=boo[ey[j]]=1;
		}
	}
	for (int i=1; i<=n; i++) if (b[i])
		for (int nw,j=i+1; j<=n; j++) if (b[j])
		{
			nw=1;
			for (int k=1; k<=n; k++)
				if (bo[i][k]&&bo[j][k]) {nw=0; break;}
			ans+=nw;
		}
	return printf("%d\n",ans),0;
}