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>

const int N = 410, M = 100010;
typedef std::bitset <N> bit;

int V, E;
int u[M], v[M];
bit G[N];

int main() {
	int i, j, ans = 0;
	scanf("%d%d", &V, &E);
	for (i = 1; i <= E; ++i) scanf("%d%d", u + i, v + i);
	for (i = 1; i <= V; ++i)
		for (G[i].set(i), j = E; j; --j) {
			if (G[i].test(u[j]) && G[i].test(v[j])) {G[i].set(0); break;}
			if (G[i].test(u[j]) || G[i].test(v[j])) G[i].set(u[j]).set(v[j]);
		}
	for (i = 1; i < V; ++i)
		if (!G[i].test(0))
			for (j = i + 1; j <= V; ++j)
				ans += !G[j].test(0) && (G[i] & G[j]).none();
	printf("%d\n", ans);
	return 0;
}