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>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 100005
#define ll long long
int n, m, f[N], cnt[3], num, flg;
int fir[N], nex[N << 1], tar[N << 1], len[N << 1], tot;
ll ans;
void Add(int a,int b,int c)
{
	tot++;
	tar[tot] = b;
	len[tot] = c;
	nex[tot] = fir[a];
	fir[a] = tot;
}
void Dfs(int u)
{
	cnt[f[u]]++;
	for (int i = fir[u]; i; i = nex[i])
	{
		num++;
		int v = tar[i];
		if (f[v] == -1)
		{
			f[v] = (f[u] + len[i]) % 3;
			Dfs(v);
		}
		else
			if ((f[u] + len[i]) % 3 != f[v])
				flg = 0;
	}
}
int main()
{
	scanf("%d%d", &n, &m);
	while (m--)
	{
		int a, b;
		scanf("%d%d", &a, &b);
		Add(a, b, 1);
		Add(b, a, 2);
	}
	memset(f, -1, sizeof(f));
	for (int i = 1; i <= n; i++)
		if (f[i] == -1)
		{
			flg = 1;
			cnt[0] = cnt[1] = cnt[2] = 0;
			num = 0;
			f[i] = 0;
			Dfs(i);
			num >>= 1;
			if (!flg)
				ans += (ll)(cnt[0] + cnt[1] + cnt[2]) * (cnt[0] + cnt[1] + cnt[2]);
			else
				if (cnt[0] && cnt[1] && cnt[2])
					ans += (ll)cnt[0] * cnt[1] + (ll)cnt[1] * cnt[2] + (ll)cnt[2] * cnt[0];
				else
					ans += num;
		}
	printf("%lld\n", ans);
}