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 = 404, M = 1e5 + 100;

int n, m, x[M], y[M];
bool bad[N];
bitset<N> s[N];

void input() {
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		cin >> x[i] >> y[i];
		x[i]--;
		y[i]--;
	}
}

void ready(int v) {
	s[v][v] = true;
	for (int i = m - 1; ~i; i--) {
		int a = x[i], b = y[i];
		if (s[v][a] && s[v][b]) {
			bad[v] = true;
			return;
		}
		if (s[v][a])
			s[v][b] = true;
		if (s[v][b])
			s[v][a] = true;
	}
}

void Ready() {
	for (int v = 0; v < n; v++)
		ready(v);
}

void output() {
	int ans = 0;
	for (int u = 0; u < n; u++)
		for (int v = u + 1; v < n; v++)
			if (!(bad[u] || bad[v] || (s[u] & s[v]).count()))
				ans++;
	cout << ans << '\n';
}

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	input();
	Ready();
	output();
}
