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>

const int maxn = 450;

const int maxm = 1e5 + 100;

int n, m, ans;
int x[maxm], y[maxm];
bool f[maxn][maxn], ok[maxn];

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