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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=100005;
int n,m,i,j,k,head[N],adj[N],nxt[N],rhead[N],radj[N],rnxt[N],p[N],cnt,col[N],w[3],num;
bool t,v[N];
long long ans;
void dfs(int x)
{
	p[++cnt]=x;
	v[x]=true;
	for(int y=head[x];y;y=nxt[y],++num)
		if(!v[adj[y]])
		{
			col[adj[y]]=(col[x]+1)%3;
			dfs(adj[y]);
		}
		else if(col[adj[y]]!=(col[x]+1)%3)
			t=false;
	for(int y=rhead[x];y;y=rnxt[y])
		if(!v[radj[y]])
		{
			col[radj[y]]=(col[x]-1+3)%3;
			dfs(radj[y]);
		}
		else if(col[radj[y]]!=(col[x]-1+3)%3)
			t=false;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;++i)
	{
		scanf("%d%d",&j,&k);
		adj[i]=k;
		nxt[i]=head[j];
		head[j]=i;
		radj[i]=j;
		rnxt[i]=rhead[k];
		rhead[k]=i;
	}
	for(i=1;i<=n;++i)
		if(!v[i])
		{
			cnt=0;
			col[i]=num=0;
			t=true;
			dfs(i);
			if(!t)
				ans+=1ll*cnt*cnt;
			else
			{
				w[0]=w[1]=w[2]=0;
				for(j=1;j<=cnt;++j)
					w[col[p[j]]]++;
				if(w[0]&&w[1]&&w[2])
					ans+=1ll*w[0]*w[1]+1ll*w[1]*w[2]+1ll*w[2]*w[0];
				else
					ans+=num;
			}
		}
	printf("%lld",ans);
	return 0;
}