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;
int n, m, p[100007][2], res;
bitset<407>is[407], no;
int read()
{
	int num = 0;
	char c = getchar();
	while (c < '0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int main()
{
	n = read();
	m = read();
	for (int i = 1; i <= m; i++)
	{
		p[i][0] = read();
		p[i][1] = read();
	}
	for (int i = 1; i <= n; i++)
	{
		is[i][i] = 1;
		for (int j = m; j >= 1; j--)
		{
			if (!is[i][p[j][0]] && !is[i][p[j][1]])continue;
			if (is[i][p[j][0]] && is[i][p[j][1]])
			{
				no[i] = 1;
				break;
			}
			if (!is[i][p[j][0]])is[i][p[j][0]] = 1;
			if (!is[i][p[j][1]])is[i][p[j][1]] = 1;
		}
	}
	for (int i = 1; i <= n; i++)if (!no[i])
		for (int j = i + 1; j <= n; j++)if (!no[j])
			if (!((is[i] & is[j]).count()))
				res++;
	cout << res << endl;
	return 0;
}