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 <cstdio>
#include <bitset>
std::bitset<405> mask[405];
int x[100005], y[100005];
bool die[405];
int main()
{
	// freopen("AGC016-E.in", "r", stdin);
	int n, m, ans = 0;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++)
	{
		scanf("%d%d", x + i, y + i);
		x[i]--;
		y[i]--;
	}
	for (int i = 0; i < n; i++)
	{
		mask[i][i] = 1;
		for (int j = m - 1; j >= 0; j--)
		{
			if (mask[i][x[j]] && mask[i][y[j]])
				die[i] = true;
			else if (mask[i][x[j]])
				mask[i][y[j]] = true;
			else if (mask[i][y[j]])
				mask[i][x[j]] = true;
		}
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = i + 1; j < n; j++)
			ans += !die[i] && !die[j] && !(mask[i] & mask[j]).count();
	}
	printf("%d\n", ans);
	return 0;
}
