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 pd[410],u[110000],v[110000],n,m;
bitset<410>A[410],x[110000],ans,y[110000];
int main()
{
	scanf("%d%d",&n,&m);
	for (int i = 1; i <= m; i ++) scanf("%d%d",&u[i],&v[i]);
	for (int i = 1; i <= n; i ++) A[i][i]=1;
	for (int i = m; i; i --)
	{
		x[i] = A[u[i]];
		y[i] = A[v[i]];
		A[u[i]] |= A[v[i]];
		A[v[i]] = A[u[i]];
	}
	int num = 0;
	for (int i = 1; i <= m; i ++)
		for (int j = 1; j <= n; j ++)
			if (x[i][j] && y[i][j]) pd[j]=1;
	for (int i = 1; i <= n; i ++)
	{
		ans = 0; 
		if (pd[i]) continue;
		for (int j = 1; j <= m; j ++)
		{
			if (x[j][i]) ans |= y[j];
			if (y[j][i]) ans |= x[j];
		}
		for (int j = i + 1; j <= n; j ++) 
			if (!ans[j] && !pd[j]) num ++;
	}
	cout << num << endl;
	return 0;
}