Problem:
Score : 1700 points

Problem Statement

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.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

3

It is possible to paint one white cell black, as follows:

Since cells (1, 2) and (2, 3) are both black and a cell (3, 1) is white, paint the cell (3, 1) black.

Sample Input 2

2 2
1 1
1 2

Sample Output 2

4

It is possible to paint two white cells black, as follows:

Since cells (1, 1) and (1, 2) are both black and a cell (2, 1) is white, paint the cell (2, 1) black.

Since cells (2, 1) and (1, 2) are both black and a cell (2, 2) is white, paint the cell (2, 2) black.

Sample Input 3

4 3
1 2
1 3
4 4

Sample Output 3

3

No white cells can be painted black.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <vector> 

typedef long long LL;
const int MN = 100005;

int N, M; LL Ans;
std::vector<int> G[MN];

int ok, col[MN], a[3], cv, ce;
void DFS(int u, int c) {
	++a[col[u] = c], ++cv; ++c %= 3;
	for (int v : G[u]) {
		++ce;
		if (!~col[v]) DFS(v, c);
		else if (col[v] != c) ok = 0;
	}
}

int main() {
	scanf("%d%d", &N, &M);
	while (M--) {
		int x, y;
		scanf("%d%d", &x, &y);
		G[x].push_back(y);
	}
	for (int i = 1; i <= N; ++i) col[i] = -1;
	for (int i = 1; i <= N; ++i) if (!~col[i]) {
		ok = 1, a[0] = a[1] = a[2] = cv = ce = 0;
		DFS(i, 0);
		if (!ok) Ans += (LL)cv * cv;
		else if (a[0] && a[1] && a[2]) Ans += (LL)a[0] * a[1] + (LL)a[1] * a[2] + (LL)a[2] * a[0];
		else Ans += ce;
	}
	printf("%lld\n", Ans);
	return 0;
}