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>
using namespace std;

bitset<400> u[400];
bitset<400> die;
int x[100005], y[100005];
int n,m;

int main()
{
  scanf("%d%d", &n, &m);  
  for (int i = 0; i < m;i++) {
    scanf("%d%d", x+i, y+i);
  }

  for (int i = 0; i < n;i++) u[i].set(i);

  for (int i = m-1; i>=0; i--) {
    int p = x[i] - 1, q = y[i] - 1;
    die |= (u[p] & u[q]);
    u[p] |= u[q];
    u[q] |= u[p];
  }

  int ans = 0;
  for (int i = 0; i < n; i++)
  for (int j = i+1; j < n; j++) if (!die.test(i) && !die.test(j)) {
    ans++;
    for (int k = 0; k < n; k++) if (u[k].test(i) && u[k].test(j)) {
      ans--;
      break;
    }
  }
  printf("%d\n", ans);
  return 0;
}