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 <iostream>
#define MN 410
#define MM 101000

int n, m;
int x[MM], y[MM];
bool used[MN][MN], ok[MN];

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 <= n; i++)
	{
		used[i][i] = 1;
		for(int j = m; j >= 1; j--)
		{
			if(used[i][x[j]] && used[i][y[j]]) {ok[i] = 1; break;}
			else if(used[i][x[j]]) used[i][y[j]] = 1;
			else if(used[i][y[j]]) used[i][x[j]] = 1; 
		}
	}
	int ans = 0;
	for(int i = 1; i <= n; i++)
		for(int j = i + 1; j <= n; j++)
		{
			if(ok[i] || ok[j]) continue;
			bool okk = 0;
			for(int k = 1; k <= n; k++)
				if(used[i][k] && used[j][k]) okk = 1;
			if(!okk) ans++;
		}
	printf("%d\n", ans);
}