Problem:
Score : 1400 points

Problem Statement

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.

Sample Input 1

3 1
1 2

Sample Output 1

2

(i,\ j) = (1,\ 3), (2,\ 3) satisfy the condition.

Sample Input 2

4 3
1 2
3 4
2 3

Sample Output 2

1

(i,\ j) = (1,\ 4) satisfies the condition.
Both turkeys 1 and 4 are alive if:

The first man eats turkey 2.

The second man eats turkey 3.

The third man does nothing.

Sample Input 3

3 2
1 2
1 2

Sample Output 3

0

Sample Input 4

10 10
8 9
2 8
4 6
4 9
7 8
2 8
1 8
3 4
3 4
2 7

Sample Output 4

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

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

bitset<400> u[400], ru[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);
    ru[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];
  }

  for (int i = 0; i < m; i++) {
    int p = x[i] - 1, q = y[i] - 1;
    ru[p] |= ru[q];
    ru[q] |= ru[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 (ru[k].test(i) && ru[k].test(j)) {
            ans--;
            break;
          }
      }
  printf("%d\n", ans);
  return 0;
}