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 MAXN = 410;
int n, m, xs[100010], ys[100010], ans;
std::bitset<MAXN> B[MAXN];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> m;
	for (int i = 1; i <= m; ++i) std::cin >> xs[i] >> ys[i];
	for (int i = 1; i <= n; ++i) {
		B[i].set(i);
		for (int j = m; j; --j)
			if (B[i][xs[j]] && B[i][ys[j]]) B[i].set();
			else if (B[i][xs[j]]) B[i].set(ys[j]);
			else if (B[i][ys[j]]) B[i].set(xs[j]);
		B[i].reset(0);
	}
	for (int i = 1; i <= n; ++i)
		for (int j = i + 1; j <= n; ++j)
			ans += (B[i] & B[j]).none();
	std::cout << ans << std::endl;
	return 0;
}
