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<cstring>
#include<cstdio>
#include<algorithm>
#define N 100010
#define lint long long
#define gc getchar()
#define debug(x) cerr<<#x<<"="<<x
#define sp <<" "
#define ln <<endl
using namespace std;
inline int inn()
{
	int x,ch;while((ch=gc)<'0'||ch>'9');
	x=ch^'0';while((ch=gc)>='0'&&ch<='9')
		x=(x<<1)+(x<<3)+(ch^'0');return x;
}
struct edges{
	int to,pre,wgt;
}e[N<<1];int h[N],etop,vis[N],col[N],can;lint nc,mc,a[5];
inline int add_edge(int u,int v,int w)
{	return e[++etop].to=v,e[etop].pre=h[u],h[u]=etop,e[etop].wgt=w;	}
inline int mol(int x) { return x%=3,(x<0?x+3:x); }
int dfs(int x)
{
	a[col[x]]++,vis[x]=1,nc++;
	for(int i=h[x],y;i;i=e[i].pre) mc+=(e[i].wgt==1);
	for(int i=h[x],y;i;i=e[i].pre)
		if(!vis[y=e[i].to]) col[y]=mol(col[x]+e[i].wgt),dfs(y);
		else if(col[y]!=mol(col[x]+e[i].wgt)) can=0;
	return 0;
}
int main()
{
	int n=inn(),m=inn();lint ans=0;
	for(int i=1,u,v;i<=m;i++)
		u=inn(),v=inn(),add_edge(u,v,1),add_edge(v,u,-1);
	for(int i=1;i<=n;i++) if(!vis[i])
	{
		can=1,mc=nc=a[0]=a[1]=a[2]=0ll,dfs(i);
//		debug(i)sp,debug(can)sp,debug(a[0])sp,debug(a[1])sp,debug(a[2])ln;
		if(!can) ans+=nc*nc;
		else if(a[0]&&a[1]&&a[2]) ans+=a[0]*a[1]+a[1]*a[2]+a[2]*a[0];
		else ans+=mc;
	}
	return !printf("%lld\n",ans);
}