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
// O(N^2 M)
#include <bits/stdc++.h>
using namespace std;

int n, m;
int a[100000];
int b[100000];
bool used[400];
bool sub(int s, int t) {
  for (int i = 0; i < n; ++i) used[i] = false;
  used[s] = true;
  used[t] = true;
  for (int j = m - 1; j >= 0; --j) {
    int x = a[j];
    int y = b[j];
    if (used[x] && used[y]) {
      return false;
    } else if (used[x]) {
      used[y] = true;
    } else if (used[y]) {
      used[x] = true;
    }
  }
  return true;
}

int main() {
  cin >> n >> m;
  for (int j = 0; j < m; ++j) {
    int x, y;
    cin >> x >> y;
    --x; --y;
    a[j] = x;
    b[j] = y;
  }
  
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      if (sub(i, j)) ++cnt;
    }
  }
  cout << cnt << endl;
  return 0;
}