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 <cstdio>

#define N 100100
#define ll long long

struct Edge {
	int v, w, nxt;
} e[N << 1];
int head[N << 1], cnt;
inline void addedge(int u, int v, int w) {
	e[++cnt] = (Edge){v, w, head[u]};
	head[u] = cnt;
}

int col[N], flag, num[4];
int dye(int x, int c) {
	if (col[x])	{
		if (col[x] != c)
			flag = false;
		return 0;
	}
	col[x] = c;
	num[c]++;
	num[0]++;
	int ans = 0;
	for (int i = head[x]; i; i = e[i].nxt)
		ans += 1 + dye(e[i].v, (c + e[i].w) % 3 + 1);
	return ans;
}

int main() {
	int n, m;
	scanf("%d %d", &n, &m);
	for (int i = 1, a, b; i <= m; i++) {
		scanf("%d %d", &a, &b);
		addedge(a, b, 0);
		addedge(b, a, 1);
	}
	ll ans = 0;
	for (int i = 1; i <= n; i++)
		if (!col[i]) {
			num[0] = num[1] = num[2] = num[3] = 0;
			flag = true;
			int ret = dye(i, 1);
			if (flag == 0)
				ans += (ll)num[0] * num[0];
			else if (num[1] && num[2] && num[3])
				ans += (ll)num[1] * num[2] + (ll)num[2] * num[3] + (ll)num[3] * num[1];
			else
				ans += ret / 2;
		}
	printf("%lld\n", ans);
}