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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define M 200020
using namespace std;
int read(){
	int nm=0,fh=1; char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) if(cw=='-') fh=-fh;
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return nm*fh;
}
int n,m,fs[M],nt[M<<1],to[M<<1],len[M<<1],tmp;
int f[M],sz[M],w[M],u,v,ct[M],cnt[4];
bool vis[M];
LL ans;
int fd(int x){return x==f[x]?x:f[x]=fd(f[x]);}
void link(int x,int y,int dst){nt[tmp]=fs[x],fs[x]=tmp,to[tmp]=y,len[tmp++]=dst;}
LL mul(int x,int y){return (LL)x*(LL)y;}
bool dfs(int x,int now){
	now+=now<0?3:0,now-=now>2?3:0;
	if(vis[x]&&now!=w[x]) return false;
	else if(vis[x]) return true;
	vis[x]=true,cnt[now]++,w[x]=now;
	for(int i=fs[x];i!=-1;i=nt[i]) if(!dfs(to[i],now+len[i])) return false;
	return true;
}
int main(){
	n=read(),m=read();
	for(int i=1;i<=n;i++) f[i]=i,sz[i]=1,fs[i]=-1,vis[i]=false;
	for(int i=1;i<=m;i++){
	    u=read(),v=read(),link(u,v,1),link(v,u,-1);
	    u=fd(u),v=fd(v),ct[v]++;
	    if(u!=v) f[u]=v,sz[v]+=sz[u],ct[v]+=ct[u];
	}
	for(int i=1;i<=n;i++){
		if(i!=fd(i)) continue;
		cnt[0]=cnt[1]=cnt[2]=0;
		if(!dfs(i,0)) ans+=mul(sz[i],sz[i]);
		else if(!cnt[0]||!cnt[1]||!cnt[2]) ans+=ct[i];
		else ans+=mul(cnt[0],cnt[1])+mul(cnt[1],cnt[2])+mul(cnt[2],cnt[0]);
	}
	printf("%lld\n",ans); return 0;
}