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>
#define ll long long
using namespace std;
int cnt,fst[200005],nxt[200005],to[200005],w[200005];
int n,m,col[200005],sum[3],res,snd;
ll ans;
bool vis[200005],fg;
void AddEdge(int u,int v,int c)
{
	to[++cnt]=v;
	nxt[cnt]=fst[u];
	fst[u]=cnt;
	w[cnt]=c;
}
int Calc(int x)
{
	return x>=3?x-3:(x<0?x+3:x);
}
void Dfs(int u)
{
	snd++;
	sum[col[u]]++;
	vis[u]=1;
	for(int i=fst[u];i;i=nxt[i])
	{
		int v=to[i];
		if(w[i]==1) res++;
		if(!vis[v])
		{
			col[v]=Calc(col[u]+w[i]);
			Dfs(v);
		}
		else if(col[v]!=Calc(col[u]+w[i])) fg=0;
	}
}
int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d %d",&x,&y);
		AddEdge(x,y,1);
		AddEdge(y,x,-1);
	}
	for(int i=1;i<=n;i++)
	{
		if(vis[i]) continue;
		memset(sum,0,sizeof(sum));
		res=snd=0;
		fg=1;
		Dfs(i);
		if(!fg) ans+=1ll*snd*snd;
		else if(sum[0] && sum[1] && sum[2]) ans+=1ll*sum[0]*sum[1]+1ll*sum[0]*sum[2]+1ll*sum[1]*sum[2];
		else ans+=1ll*res;
	}
	printf("%lld\n",ans);
	return 0;
}