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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define N 110000
#define LL long long
using namespace std;
struct node{int y,nex,o;}a[2*N];
int n,m,len,fir[N],c[N];
LL ans,d[5],cnt,tmp;
bool v[N],bo;
void ins(int x,int y)
{
	a[++len].y=y;a[len].o=1;a[len].nex=fir[x];fir[x]=len;
	a[++len].y=x;a[len].o=-1;a[len].nex=fir[y];fir[y]=len;
}
void dfs(int x)
{
	cnt++;
	v[x]=1;
	d[c[x]]++;
	for(int k=fir[x];k;k=a[k].nex)
	{
		tmp++;
		int y=a[k].y,t=(c[x]+a[k].o+3)%3;
		if(v[y]) {if(c[y]!=t) bo=1;}
		else {c[y]=t;dfs(y);}
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		ins(x,y);
	}
	for(int i=1;i<=n;i++)
	{
		if(v[i]) continue;
		bo=d[0]=d[1]=d[2]=cnt=tmp=0;
		dfs(i);
		if(bo==0)
		{
			if(d[0]*d[1]*d[2]==0) ans+=tmp/2;
			else ans+=d[0]*d[1]+d[1]*d[2]+d[2]*d[0];
		}
		else ans+=cnt*cnt;
	}
	printf("%lld\n",ans);
	return 0;
}