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>
int N, M, u[100001], v[100001], disable[401], O;
std::bitset < 401 > mask[401];
int main()
{
	scanf("%d%d", &N, &M);
	for (int i = 1; i <= M; i++)
		scanf("%d%d", u + i, v + i);
	for (int i = 1; i <= N; i++)
	{
		std::bitset < 401 > &p = mask[i];
		p.set(i);
		for (int j = M; j; j--)
			if (p[u[j]] && p[v[j]])
			{
				disable[i] = 1;
				break;
			}
			else if (p[u[j]])
				p.set(v[j]);
			else if (p[v[j]])
				p.set(u[j]);
	}
	for (int i = 1; i < N; i++)
		if (!disable[i])
			for (int j = i + 1; j <= N; j++)
				if (!disable[j] && !(mask[i] & mask[j]).count())
					O++;
	printf("%d\n", O);
	return 0;
}
