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 nn 65536*3
#define ll long long
int pa[nn],siz[nn];
int n,m;
vector<int> v[nn],c[nn],rev[nn];
int modi(int x)
{
	x+=3;return x%3;
}
void add_edge(int x,int y)
{
	v[x].push_back(y);v[y].push_back(x);
	c[x].push_back(1);c[y].push_back(-1);
}


int cnt[4];int fail;int vis[nn];int col[nn];
void dfs(int x,int cc)
{
	vis[x]=1;col[x]=cc;cnt[cc]++;
	for(int i=0;i<v[x].size();i++)
	{
		int y=v[x][i];
		if(c[x][i]==1) cnt[3]++;
		if(!vis[y]) dfs(y,modi(cc+c[x][i]));
		else if(col[y]!=modi(cc+c[x][i])) fail=1;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	while(m--)
	{
		int x,y;scanf("%d%d",&x,&y);
		add_edge(x,y);
	}
	ll ans=0;
	for(int i=1;i<=n;i++)
	{
		if(!vis[i]) 
		{
			cnt[0]=cnt[1]=cnt[2]=cnt[3]=0;fail=0;
			dfs(i,0);int tot=cnt[0]+cnt[1]+cnt[2];
			
			if(fail==1) ans+=1ll*tot*tot;
			else if(!cnt[0] or !cnt[1] or !cnt[2]) ans+=cnt[3];
			else ans+=1ll*cnt[0]*cnt[1]+1ll*cnt[1]*cnt[2]+1ll*cnt[2]*cnt[0];
		}
	}
	cout<<ans;
	return 0;
}