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;
#define LL long long

const int maxn=2e5+10;

int n,m,col[maxn],sz[maxn],c[9],edge_cnt;
int tote,NEXT[maxn],TO[maxn],FIR[maxn],INV[maxn];
LL Ans;

void addedge(int G[],int u,int v)
{
	TO[++tote]=v;
	NEXT[tote]=G[u];
	G[u]=tote;
}

bool dfs(int u)
{
	bool Ans=1;
	c[col[u]]++;
	sz[u]=1;
	for (int p=FIR[u];p;p=NEXT[p])
	{
		edge_cnt++;
		int v=TO[p];
		if (!col[v])
		{
			col[v]=col[u]%3+1;
			Ans&=dfs(v);
			sz[u]+=sz[v];
		}
		else Ans&=col[v]==(col[u]%3+1);
	}
	for (int p=INV[u];p;p=NEXT[p])
	{
		edge_cnt++;
		int v=TO[p];
		if (!col[v])
		{
			col[v]=(col[u]+1)%3+1;
			Ans&=dfs(v);
			sz[u]+=sz[v];
		}
		else Ans&=col[v]==((col[u]+1)%3+1);
	}
	return Ans;
}

int main()
{
#ifdef h10
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
#endif
	int i,u,v;
	scanf("%d%d",&n,&m);
	for (i=1;i<=m;i++)
	{
		scanf("%d%d",&u,&v);
		addedge(FIR,u,v);
		addedge(INV,v,u);
	}
	for (i=1;i<=n;i++)
		if (!col[i])
		{
			edge_cnt=0;
			memset(c,0,sizeof(c));
			col[i]=1;
			if (dfs(i))
			{
				if (c[1]&&c[2]&&c[3])
				{
					Ans+=(LL)c[1]*c[2];
					Ans+=(LL)c[2]*c[3];
					Ans+=(LL)c[3]*c[1];
				}
				else Ans+=edge_cnt>>1;
			}
			else Ans+=(LL)sz[i]*sz[i];
		}
	printf("%lld\n",Ans);
}