Problem: We have a grid with N rows and N columns. The cell at the i-th row and j-th column is denoted (i, j).

Initially, M of the cells are painted black, and all other cells are white. Specifically, the cells (a_1, b_1), (a_2, b_2), ..., (a_M, b_M) are painted black.

Snuke will try to paint as many white cells black as possible, according to the following rule:

* If two cells (x, y) and (y, z) are both black and a cell (z, x) is white for integers 1≤x,y,z≤N, paint the cell (z, x) black.



Find the number of black cells when no more white cells can be painted black.

Constraints

* 1≤N≤10^5
* 1≤M≤10^5
* 1≤a_i,b_i≤N
* All pairs (a_i, b_i) are distinct.

Input

The input is given from Standard Input in the following format:


N M
a_1 b_1
a_2 b_2
:
a_M b_M


Output

Print the number of black cells when no more white cells can be painted black.

Examples

Input

3 2
1 2
2 3


Output

3


Input

2 2
1 1
1 2


Output

4


Input

4 3
1 2
1 3
4 4


Output

3

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;

const int N = 123456;

int n, m, color[N], number[3], number_e;
vector<pair<int, int>> adj[N];

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  long long answer = 0;
  for (int i = 1, x, y; i <= m; ++i) {
    cin >> x >> y;
    adj[x].emplace_back(y, 1);
    adj[y].emplace_back(x, -1);
  }
  bool ok;
  function<void (int, int)> dfs = [&] (int x, int z) {
    ++number[color[x] = z];
    for (auto e : adj[x]) {
      int y = e.first;
      if (e.second == 1) {
        ++number_e;
      }
      if (!~color[y]) {
        dfs(y, (z + e.second + 3) % 3);
      } else if (color[y] != (z + e.second + 3) % 3) {
        ok = false;
      }
    }
  };
  memset(color, -1, sizeof color);
  for (int i = 1; i <= n; ++i) {
    if (!~color[i]) {
      memset(number, 0, sizeof number);
      number_e = 0;
      ok = true;
      dfs(i, 0);
      if (ok && (!number[0] || !number[1] || !number[2])) {
        answer += number_e;
      } else if (ok) {
        answer += (long long) number[0] * number[1];
        answer += (long long) number[1] * number[2];
        answer += (long long) number[2] * number[0];
      } else {
        int all = number[0] + number[1] + number[2];
        answer += (long long) all * all;
      }
    }
  }
  cout << answer << '\n';
  return 0;
}