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
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;

typedef long long LL;
const int N = 1e5 + 50;
const int incol[] = { 1, 2, 0 };
const int outcol[] = { 2, 0, 1 };

struct edge {
  int nxt, to;
} e[N << 1];
int fir[N], cnt = 0;
int n, m, col[N], tms[4], ce;
bool bad;

inline void addedge(int x, int y) {
  e[++ cnt] = (edge) { fir[x], y }; fir[x] = cnt;
}

inline void Dfs(int x) {
  int exp;
  tms[col[x]] ++;
  for (int i = fir[x]; i; i = e[i].nxt) {
    ce ++;
    if (i & 1) exp = outcol[col[x]];
    else exp = incol[col[x]];
    if (~ col[e[i].to]) {
      if (exp != col[e[i].to]) bad = 1;
    }
    else {
      col[e[i].to] = exp; Dfs(e[i].to);
    }
  }
}

int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, x, y; i <= m; i ++) {
    scanf("%d%d", &x, &y);
    addedge(x, y); addedge(y, x);
  }
  memset(col, -1, sizeof col);
  LL ans = 0;
  for (int i = 1; i <= n; i ++)
    if (col[i] == -1) {
      ce = 0;
      tms[0] = tms[1] = tms[2] = 0;
      bad = 0; col[i] = 0; Dfs(i);
      if (bad) {
        int sum = tms[0] + tms[1] + tms[2];
        ans += (LL) sum * sum;
      }
      else if (tms[0] && tms[1] && tms[2])
        ans += (LL) tms[0] * tms[1] + (LL) tms[0] * tms[2] + (LL) tms[1] * tms[2];
      else ans += ce >> 1;
    }
  printf("%lld\n", ans);
  return 0;
}
