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;
const int N = 410, M = 100010;
int n, m, X[M], Y[M], ans;
bool k[N][N], vis[N];

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++) {
		k[i][i] = 1;
		for(int j = m; j; j--) {
			bool x = k[i][X[j]], y = k[i][Y[j]];
			if(x && y) { vis[i] = 1; break; }
			else if(x) k[i][Y[j]] = 1; else if(y) k[i][X[j]] = 1;
		}
	}
	for(int i = 1; i < n; i++) {
		if(vis[i]) continue;
		for(int j = i + 1; j <= n; j++) {
			if(vis[j]) continue;
			bool flag = 1;
			for(int h = 1; h <= n; h++)
				if(k[i][h] && k[j][h]) { flag = 0; break; }
			ans += flag;
		}
	}
	printf("%d\n", ans);
	return 0;
}