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 <iostream>
#include <bitset>
using namespace std;

const int N = 404;
const int M = 1e5 + 50;

bitset <N> tur[N], res;
int x[M], y[M], n, m;
bool die[N];

int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i ++) tur[i][i] = 1;
  for (int i = 1; i <= m; i ++) scanf("%d%d", x + i, y + i);
  for (register int j = m, a, b; j; j --) {
    a = x[j]; b = y[j];
    for (register int i = 1; i <= n; i ++) {
      if (tur[i][a] && tur[i][b]) die[i] = 1;
      else if (tur[i][a] ^ tur[i][b]) tur[i][a] = tur[i][b] = 1;
    }
  }
  int cnt = 0;
  for (int i = 1; i <= n; i ++)
    if (!die[i])
      for (int j = i + 1; j <= n; j ++)
        if (!die[j]) {
          res = tur[i] & tur[j];
          if (!res.count()) cnt ++;
        }
  printf("%d\n", cnt);
  return 0;
}
