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>
const int N=1e5+7;
int n,m,pv[4]={0,3,1,2},nx[4]={0,2,3,1},t[4],cn,cm,sl;
std::vector<int>e[N],er[N];
int col[N];
long long ans=0;
void dfs(int w,int c){
	if(col[w]){
		if(col[w]!=c)sl=1;
		return;
	}
	++cn;
	++t[col[w]=c];
	cm+=e[w].size();
	for(int i=0;i<e[w].size();++i)dfs(e[w][i],nx[c]);
	for(int i=0;i<er[w].size();++i)dfs(er[w][i],pv[c]);
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1,a,b;i<=m;++i){
		scanf("%d%d",&a,&b);
		e[a].push_back(b);
		er[b].push_back(a);
	}
	for(int i=1;i<=n;++i)if(!col[i]){
		t[1]=t[2]=t[3]=0;
		cn=cm=sl=0;
		dfs(i,1);
		if(sl)ans+=1ll*cn*cn;
		else if(t[1]&&t[2]&&t[3])ans+=1ll*t[1]*t[2]+1ll*t[2]*t[3]+1ll*t[3]*t[1];
		else ans+=cm;
	}
	printf("%lld\n",ans);
	return 0;
}