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>
#define int long long
using namespace std;
const int N=100005;
int n,m,u,v,cnt,tot,ans,head[N],ck[N],c[3],to[N*2],nxt[N*2],dd[N*2];
bool flag;
void adde(int u,int v,int d){
	to[++cnt]=v;
	nxt[cnt]=head[u];
	dd[cnt]=d;
	head[u]=cnt;
}
void dfs(int u){
	c[ck[u]]++;
	for(int i=head[u];i;i=nxt[i]){
		if(dd[i]==1){
			tot++;
		}
		int v=to[i];
		if(ck[v]!=-1){
			if((ck[u]+dd[i])%3!=ck[v]){
				flag=true;
			}
		}else{
			ck[v]=(ck[u]+dd[i])%3;
			dfs(v);
		}
	}
}
signed main(){
	scanf("%lld%lld",&n,&m);
	for(int i=1;i<=m;i++){
		scanf("%lld%lld",&u,&v);
		adde(u,v,1);
		adde(v,u,2);
	}
	memset(ck,-1,sizeof(ck));
	for(int i=1;i<=n;i++){
		if(ck[i]==-1){
			ck[i]=tot=c[0]=c[1]=c[2]=flag=0;
			dfs(i);
			if(flag){
				ans+=(c[0]+c[1]+c[2])*(c[0]+c[1]+c[2]);
			}else if(!c[0]||!c[1]||!c[2]){
				ans+=tot;
			}else{
				ans+=c[0]*c[1]+c[1]*c[2]+c[2]*c[0];
			}
		}
	}
	printf("%lld\n",ans);
	return 0;
}