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 ref(i,x,y)for(int i=x;i<=y;++i)
const int N=100010;
int n,m,fg,d[N],t[3],S,w;long long ans;vector<pair<int,int> > e[N];
void dfs(int x){
	t[d[x]]++;S++;
	ref(i,0,((int)e[x].size())-1){
		int y=e[x][i].first;w++;
		int D=(3+d[x]+e[x][i].second)%3;
		if(d[y]==-1)d[y]=D,dfs(y);else if(d[y]!=D)fg=1;
	}
}
int main(){
	cin>>n>>m;
	ref(i,1,m){int x,y;cin>>x>>y;
		e[x].push_back(make_pair(y,1));
		e[y].push_back(make_pair(x,-1));
	}
	ref(i,1,n)d[i]=-1;
	ref(i,1,n)if(d[i]<0){
		fg=d[i]=S=w=t[0]=t[1]=t[2]=0,dfs(i);
		if(fg)ans+=1LL*S*S;else
		if(!t[0]||!t[1]||!t[2])ans+=w/2;else
		ans+=1LL*t[0]*t[1]+1LL*t[1]*t[2]+1LL*t[2]*t[0];
	}
	cout<<ans<<endl;
}