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;
const int N=2e5+10;
int siz,he[N],e,edg,col[N];
bool vis[N],flag;
struct sa{
	int ne,va,to;
} a[N];
long long ans=0,cot[3];
void add(int u,int v,int w){
	a[++e].to=v;
	a[e].va=w;
	a[e].ne=he[u];
	he[u]=e;
}
void dfs(int k){
	cot[col[k]]++;
	vis[k]=1;
	siz++;
	for(int i=he[k];i;i=a[i].ne){
		edg+=(a[i].va==1);
		int v=a[i].to;
		int rec=(3+col[k]+a[i].va)%3;
		if(!vis[v]){
		    col[v]=rec;
		    dfs(v);
	    }
	    else if(col[v]!=rec) flag=0;
	}
}
int main(){
	int n,m;
	cin>>n>>m;
	for(int i=1,u,v;i<=m;i++){
		scanf("%d%d",&u,&v);
		add(u,v,1);
		add(v,u,-1);
	}
	for(int i=1;i<=n;i++) if(!vis[i]){
		cot[1]=cot[0]=cot[2]=0;
		flag=1;
		edg=0;
		siz=0;
		dfs(i);
		if(!flag) ans+=(long long)siz*siz;
		else if(cot[0]&&cot[1]&&cot[2]) ans+=(long long) cot[0]*cot[1]+(long long) cot[0]*cot[2]+(long long) cot[1]*cot[2]; 
		else ans+=edg;
	}
	cout<<ans<<endl;
}