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<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
#include<vector>
#define L long long
using namespace std;
int n,m,x[100010],f[10],q,r;
vector<int> a[100010],b[100010];
long long p;
inline void dfs(int i)
{
	int j;
	f[x[i]]++;
	q++;
	for(j=0;j<a[i].size();j++)
	  {
	   r++;
	   if(!x[a[i][j]])
	     {
	      x[a[i][j]]=x[i]%3+1;
	      dfs(a[i][j]);
	     }
	   else if(x[a[i][j]]!=x[i]%3+1)
	     f[0]=1;
	  }
	for(j=0;j<b[i].size();j++)
	  {
	   r++;
	   if(!x[b[i][j]])
	     {
	      x[b[i][j]]=(x[i]+1)%3+1;
	      dfs(b[i][j]);
	     }
	   else if(x[b[i][j]]!=(x[i]+1)%3+1)
	     f[0]=1;
	  }
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k;
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;i++)
	  {
	   scanf("%d%d",&j,&k);
	   a[j].push_back(k);
	   b[k].push_back(j);
	  }
	for(i=1;i<=n;i++)
	  if(!x[i])
	    {
	     for(j=0;j<=3;j++)
	       f[j]=0;
	     q=r=0;
	     x[i]=1;
	     dfs(i);
	     if(f[0])
	       p+=(L)q*q;
	     else if(f[1] && f[2] && f[3])
	       p+=(L)f[1]*f[2]+(L)f[2]*f[3]+(L)f[3]*f[1];
	     else
	       p+=r/2;
	    }
	printf("%lld\n",p);
	return 0;
}