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>
using namespace std;

typedef long long  LL;
struct edge{int s,t,d,n;}e[200010];
int n,m,h[100010],col[100010],s0,s1,s2,error,tot;
LL ans;

void dfs(int x)
{
	col[x]?(col[x]==1?s1++:s2++):s0++;
	for (int i=h[x],y; y=e[i].t,i; i=e[i].n,tot++)
		if (col[y]==-1)  col[y]=(col[x]+e[i].d)%3,dfs(y);
		else  if (col[y]!=(col[x]+e[i].d)%3)  error=1;
}

void work()
{
	scanf("%d %d",&n,&m);
	for (int i=1,x,y; i<=m; i++)
		{
			scanf("%d %d",&x,&y);
			e[++tot]=(edge){x,y,1,h[x]},h[x]=tot;
			e[++tot]=(edge){y,x,2,h[y]},h[y]=tot;
		}
	memset(col,-1,sizeof(col));
	for (int i=1; i<=n; i++)
		if (col[i]==-1)
			{
				s0=s1=s2=error=tot=0,col[i]=0,dfs(i);
				if (error)  ans+=1LL*(s0+s1+s2)*(s0+s1+s2);
				else  if ((!s1)||(!s2))  ans+=tot/2;
				else  ans+=1LL*s0*s1+1LL*s0*s2+1LL*s1*s2;
			}
	printf("%lld",ans);
}

int main()
{
	work();
	return 0;
}