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 = 200010;

struct Edge {
  int to, capa, next;
} e[N];

int h[N], sum = 0, n, m;
bool vis[N];
int cmt, cnt, col[N], c[3];
bool falun;

inline int color(const int &x) {
  if (x >= 3) return x - 3;
  if (x < 0) return x + 3;
  return x;
}

void add_edge(int u, int v, int w) {
  e[++sum].to = v;
  e[sum].capa = w;
  e[sum].next = h[u];
  h[u] = sum;
}

void dfs(int u) {
  c[col[u]]++;
  vis[u] = 1;
  cnt++;
  for (int tmp = h[u]; tmp; tmp = e[tmp].next) {
    cmt += (e[tmp].capa == 1);
    int v = e[tmp].to;
    if (!vis[v])
      col[v] = color(col[u] + e[tmp].capa), dfs(v);
    else if (col[v] != color(col[u] + e[tmp].capa))
      falun = 0;
  }
}

int main() {
  int u, v;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v, 1);
    add_edge(v, u, -1);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    memset(c, 0, sizeof(c));
    cmt = cnt = 0;
    falun = 1;
    dfs(i);
    if (!falun)
      ans += 1ll * cnt * cnt;
    else if (c[0] && c[1] && c[2])
      ans += 1ll * c[0] * c[1] + 1ll * c[1] * c[2] + 1ll * c[2] * c[0];
    else
      ans += cmt;
  }
  printf("%lld\n", ans);
  return 0;
}