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 ll long long
#define rgt register
using namespace std;
const int N=200007;
struct Edge{
	int to,cost,nxt;
	Edge(int a,int b,int c):to(a),nxt(b),cost(c){ }
	Edge() { }
}b[N];
int n,m,t,v[N],head[N];
bool vis[N],r;
ll ans,res,sum,c[5];
inline int read() {
	rgt int s=0;
	rgt char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) s=(s<<1)+(s<<3)+c-'0',c=getchar();
	return s;
}
inline void add(rgt int x,rgt int y,rgt int cost) {
	b[++t]=Edge(y,head[x],cost),head[x]=t;
}
inline int Imp(rgt int x) {return (x>=3)?x-3:((x<0)?x+3:x);}
inline void Dfs(rgt int k) {
	rgt int i,to;
	++c[v[k]],++res,vis[k]=1;
	for(i=head[k];i;i=b[i].nxt) {
		to=b[i].to;
		if(b[i].cost==1) sum++;
		if(!vis[to]) v[to]=Imp(v[k]+b[i].cost),Dfs(to);
		else if(v[to]!=Imp(v[k]+b[i].cost)) r=1;
	}
}
int main()
{
	int i,x,y,cost;
	n=read(),m=read();
	for(i=1;i<=m;++i)
		x=read(),y=read(),add(x,y,1),add(y,x,-1);
	for(i=1;i<=n;++i) {
		if(vis[i]) continue;
		c[0]=c[1]=c[2]=0;
		res=sum=r=0;
		Dfs(i);
		if(r) ans+=res*res;
		else if(c[0]&&c[1]&&c[2]) ans+=c[0]*c[1]+c[0]*c[2]+c[1]*c[2];
		else ans+=sum;
	}
	printf("%lld",ans);
	return 0;
}