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>

const int N = 400 + 10;

int n, m;
int x[N * N], y[N * N];

std::bitset<N> mask[N];

std::bitset<N> solve(int s) {
  std::bitset<N> res;
  res[s] = 1;
  for (int i = m; i > 0; --i) {
    int a = x[i], b = y[i];
    if (res[a] && res[b]) {
      res.set();
      return res;
    }
    if (res[b]) std::swap(a, b);
    if (res[a]) res[b] = 1;
  }
  return res;
}

int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) scanf("%d%d", &x[i], &y[i]);
  for (int i = 1; i <= n; ++i) mask[i] = solve(i);
  int ans = 0;
  for (int i = 1; i <= n; ++i)
      for (int j = 1; j < i; ++j)
        ans += !mask[i].all() && !mask[j].all() && (mask[i] & mask[j]).none();
  printf("%d\n", ans);
  return 0;
}
