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>
int n, m, a[100009], b[100009]; bool r[409][409], f[409];
int main() {
	scanf("%d %d", &n, &m);
	for (int i = m - 1; i >= 0; i--) scanf("%d %d", &a[i], &b[i]), a[i]--, b[i]--;
	for (int i = 0; i < n; i++) {
		r[i][i] = true;
		for (int j = 0; j < m; j++) {
			if (r[i][a[j]] && r[i][b[j]]) f[i] = true;
			if (r[i][a[j]]) r[i][b[j]] = true;
			if (r[i][b[j]]) r[i][a[j]] = true;
		}
	}
	int ret = 0;
	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j < n; j++) {
			bool ok = true;
			for (int k = 0; k < n; k++) {
				if (r[i][k] && r[j][k]) ok = false;
			}
			if (ok && !f[i] && !f[j]) ret++;
		}
	}
	printf("%d\n", ret);
	return 0;
}