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>
#define N 100005
#define ll long long
using namespace std;
int n,m,sum,flag,s[4],id[N],gg;ll Ans;
vector<int>a[N],b[N];
void dfs(int x,int y)
{
	if (!id[x]) sum++,id[x]=y;
	else
	{
		if (id[x]!=y) flag=1;
		return;
	}
	s[id[x]]++;
	for (int i=0;i<a[x].size();i++)
		dfs(a[x][i],y%3+1),gg++;
	for (int i=0;i<b[x].size();i++)
		dfs(b[x][i],(y+1)%3+1),gg++;
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		a[x].push_back(y);
		b[y].push_back(x);
	}
	for (int i=1;i<=n;i++)
	{
		if (!id[i])
		{
			flag=0;sum=0;gg=0;
			s[1]=s[2]=s[3]=0;
			dfs(i,1);
			if (flag) Ans+=(ll)sum*sum;
			else if (!s[1]||!s[2]||!s[3]) Ans+=gg/2;
			else Ans+=(ll)s[1]*s[2]+(ll)s[2]*s[3]+(ll)s[3]*s[1];
		}
	}
	printf("%lld\n",Ans);	
}