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<cstdio>
#include<cstring>
#define ll long long
#define M 100010
using namespace std;
struct point{
	int to,next,f;
}e[M<<1];
int n,m,num;
ll tot1,tot2,ans;
int head[M],co[M];
ll cnt[5];
bool flag;
bool vis[M];
void add(int from,int to,int f)
{
	e[++num].next=head[from];
	e[num].to=to;
	e[num].f=f;
	head[from]=num;
}
void dfs(int x)
{
	vis[x]=true; tot1++;
	cnt[co[x]]++;
	for(int i=head[x];i;i=e[i].next)
	{
		int to=e[i].to;
		tot2+=e[i].f==1;
		if(!vis[to])
		{
			co[to]=(co[x]+e[i].f+3)%3;
			dfs(to);
		}
		else if(co[to]!=(co[x]+e[i].f+3)%3) flag=true;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y; scanf("%d%d",&x,&y);
		add(x,y,1); add(y,x,-1);
	}
	for(int i=1;i<=n;i++)
	{
		if(!vis[i])
		{
			flag=false;
			tot1=tot2=0;
			cnt[0]=cnt[1]=cnt[2]=0;
			dfs(i);
			if(flag) ans+=tot1*tot1;
			else if(cnt[0]&&cnt[1]&&cnt[2]) ans+=cnt[0]*cnt[1]+cnt[1]*cnt[2]+cnt[2]*cnt[0];
			else ans+=tot2;
		}
	}
	printf("%lld",ans);
	return 0;
}