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 MAXN = 405, MAXM = 100005;

int n, m, u[MAXM], v[MAXM];
bitset<MAXN> vis[MAXN];
bool alive[MAXN];

inline int read()
{
	int a = 0, c;
	do c = getchar(); while (!isdigit(c));
	while (isdigit(c)) {
		a = a*10+c-'0';
		c = getchar();
	}
	return a;
}

int main()
{
	n = read(), m = read();
	for (int i = 1; i <= m; i++) u[i] = read(), v[i] = read();
	for (int i = 1; i <= n; i++) {
		vis[i][i] = 1;
		alive[i] = 1;
		for (int j = m; j >= 1; j--) {
			if (vis[i][u[j]] && vis[i][v[j]]) { alive[i] = 0; break; }
			else if (vis[i][u[j]]) vis[i][v[j]] = 1;
			else if (vis[i][v[j]]) vis[i][u[j]] = 1;
		}
	}
	int cnt = 0;
	for (int i = 1; i <= n; i++)
		for (int j = i+1; j <= n; j++)
			if (alive[i] && alive[j] && (vis[i]&vis[j]).none())
				cnt++;
	cout << cnt << endl;
	return 0;
}
