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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;

const int N = 4e2 + 8;
int n, m, ans;
bool dead[N];
bool tar[N][N];

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		a--, b--;

		if (dead[a] || dead[b] || tar[a][b]) {
			dead[a] = dead[b] = true;
			continue;
		}
		
		tar[a][b] = tar[b][a] = true;
		for (int c = 0; c < n; c++)
			if (c != a && tar[c][b])
				tar[a][c] = tar[c][a] = true;
		for (int c = 0; c < n; c++)
			if (c != b && tar[c][a])
				tar[b][c] = tar[c][b] = true;
	}

	for (int i = 0; i < n; i++)
		for (int j = i + 1; j < n; j++)
			ans += !dead[i] && !dead[j] && !tar[i][j];

	cout << ans << "\n";
	return 0;
}