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>

const int MaxN = 400 + 5, MaxM = 100000 + 5;

int N, M;
int X[MaxM], Y[MaxM];
bool S[MaxN][MaxN], O[MaxN];

void init() {
  scanf("%d %d", &N, &M);
  for (int i = 1; i <= M; ++i)
    scanf("%d %d", &X[i], &Y[i]);
}

void solve() {
  for (int i = 1; i <= N; ++i) {
    S[i][i] = true; O[i] = true;
    for (int j = M; j >= 1; --j) {
      int x = X[j], y = Y[j];
      if (S[i][x] == true && S[i][y] == true) {
        O[i] = false;
        break;
      } else if (S[i][x] == true) S[i][y] = true;
      else if (S[i][y] == true) S[i][x] = true;
    }
  }

  int Ans = 0;
  for (int i = 1; i < N; ++i) for (int j = i + 1; j <= N; ++j) {
    if (O[i] == false || O[j] == false) continue;
    int add = 1;
    for (int k = 1; k <= N; ++k)
      if (S[i][k] == true && S[j][k] == true) add = 0;
    Ans += add;
  }
  printf("%d\n", Ans);
}

int main() {
  init();
  solve();
  return 0;
}