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,x[100010],y[100010],gg[405],ans;
bitset <405> a[405];

void work()
{
	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++)
		{
			a[i][i]=1;
			for (int j=m; j; j--)
				if ((a[i][x[j]])&&(a[i][y[j]]))  gg[i]=1;
				else  if ((a[i][x[j]])||(a[i][y[j]]))  a[i][x[j]]=1,a[i][y[j]]=1;
		}
	for (int i=1; i<=n; i++)
		for (int j=1; j<i; j++)
			ans+=(!gg[i])&&(!gg[j])&&(!(a[i]&a[j]).count());
	printf("%d",ans);
}

int main()
{
	work();
	return 0;
}
