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 <stdio.h>

int n, m, x, y, ans;
bool vis[410][410], dead[410];
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++) {
		scanf("%d%d", &x, &y);
		if (vis[x][y]) dead[x] = dead[y] = 1;
		if (dead[x]) dead[y] = 1;
		if (dead[y]) dead[x] = 1;
		vis[y][x] = vis[x][y] = 1;
		for (int i = 1; i <= n; i++) {
			if (i == x || i == y) continue;
			if (vis[i][x]) vis[i][y] = vis[y][i] = 1;
			if (vis[i][y]) vis[i][x] = vis[x][i] = 1;
		}
	}
	for (int i = 1; i < n; i++)
		for (int j = i + 1; j <= n; j++)
			if (!dead[i] && !dead[j] && !vis[i][j]) ans++;
	printf("%d\n", ans);
	return 0;
}