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 <iostream>
#include <cstdio>
#include <cstring>
#define N 100005
using namespace std;
typedef long long ll;
struct edge{
	int k,next;
}e[N<<2];
int n,m,home[N],cnt=-1,H[N];
void add(int *home,int x,int y){
	cnt++;
	e[cnt].k=y;
	e[cnt].next=home[x];
	home[x]=cnt;
}
int col[N],D,sum[5],P,E;
bool vis[N];
void dfs(int k){
	sum[col[k]]++;
	int w=col[k]%3+1;
	vis[k]=1;P++;
	for(int i=home[k];~i;i=e[i].next){
		E++;
		if(col[e[i].k]&&col[e[i].k]!=w){
			D=0;
		}else{
			col[e[i].k]=w;
		}
	}
	int W=col[k]-1;
	if(W==0) W=3;
	for(int i=H[k];~i;i=e[i].next) if(!vis[e[i].k]){
		if(!col[e[i].k]) col[e[i].k]=W;
		dfs(e[i].k);
	}
}
int main(){
	memset(home,-1,sizeof(home));
	memset(H,-1,sizeof(H));
	scanf("%d%d",&n,&m);
	for(int i=1,x,y;i<=m;i++){
		scanf("%d%d",&x,&y);
		add(home,x,y);
		add(H,x,y);add(H,y,x);
	}
	ll ans=0;
	for(int i=1;i<=n;i++){
		if(!col[i]){
			sum[1]=sum[2]=sum[3]=E=P=0;
			D=1,col[i]=1,dfs(i);
			if(!D) ans+=1LL*P*(P+1)-P;
			else if(!sum[3]||!sum[2]||!sum[1]) ans+=E;
			else ans+=1LL*sum[1]*sum[2]+1LL*sum[2]*sum[3]+1LL*sum[3]*sum[1];
		}
	}
	printf("%lld",ans);
	return 0;
}