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 <iostream>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>

int main() {
	int n, m;
	std::cin >> n >> m;

	static bool g[400][400];

	while (m--) {
		int x, y;
		scanf("%d %d", &x, &y);
		x--;
		y--;
		if (!g[x][y]) {
			for (int i = 0; i < n; i++) {
				if (g[x][i]) {
					g[y][i] = true;
					g[i][y] = true;
				}
				if (g[y][i]) {
					g[x][i] = true;
					g[i][x] = true;
				}
			}
		} else {
			for (int i = 0; i < n; i++) {
				g[i][x] = true;
				g[x][i] = true;
				g[i][y] = true;
				g[y][i] = true;
			}
		}
		g[x][y] = true;
		g[y][x] = true;
	}

	int ans = 0;
	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j < n; j++) {
			if (!g[i][j]) {
				ans++;
			}
		}
	}
	std::cout << ans << std::endl;
}
