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>

const int N = 400, M = 100000;

std::bitset <N> s[N];
int x[M], y[M], n, m, ans;

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n >> m;
	for (int i = 0; i < m; ++i)
		std::cin >> x[i] >> y[i], --x[i], --y[i];
	for (int i = 0; i < n; ++i) {
		s[i].set(i);
		for (int j = m - 1; ~j; --j) {
			bool a = s[i].test(x[j]), b = s[i].test(y[j]);
			if (a ^ b) s[i].set(x[j]), s[i].set(y[j]);
			else if (a && b) {s[i].set(); break;}
		}
	}
	for (int i = 0; i < n; ++i)
		for (int j = i + 1; j < n; ++j)
			ans += !(s[i] & s[j]).any();
	std::cout << ans << '\n';
	return 0;
}