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, Ans;
bitset<555> G[555], die, all;

int main(void) {
  scanf("%d%d", &n, &m);
  for(int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    G[x] |= G[y];
    G[y] |= G[x];
    for(int j = 1; j <= n; j++) {
      if(G[j].test(x)) G[j].set(y);
      if(G[j].test(y)) G[j].set(x);
    }
    G[x].set(y);
    G[y].set(x);
  }
  for(int i = 1; i <= n; i++) {
    if(G[i].test(i))
      die[i] = 1;
    all.set(i);
  }
  for(int i = 1; i <= n; i++) if(!die.test(i)) {
    G[i].flip();
    Ans = Ans + (G[i] & (G[i] ^ die) & all).count();
  }
  cout << (Ans - n + die.count()) / 2 << endl;
}
