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>
using namespace std;
 
int n,m,x,y;
int f[5],col[100005];
int tot=0,h[100005];
bool flag;
struct Edge{
	int x,dis,next;
}e[200005];
long long ans=0;
 
inline void add_edge(int x,int y,int z){
	e[++tot].x=y,e[tot].dis=z;
	e[tot].next=h[x],h[x]=tot;
}
 
void dfs(int x){
	f[col[x]]++;
	for(int i=h[x];i;i=e[i].next){
		f[3]+=e[i].dis==1;
		if(col[e[i].x]==-1){
			col[e[i].x]=(col[x]+e[i].dis)%3;
			dfs(e[i].x);
		}else{
			if(col[e[i].x]!=(col[x]+e[i].dis)%3)
				flag=true;
		}
	}
}
 
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		add_edge(x,y,1);
		add_edge(y,x,2);
	}
	memset(col,-1,sizeof(col));
	for(int i=1;i<=n;i++)
	if(col[i]==-1){
		f[col[i]=0]=f[1]=f[2]=f[3]=flag=0;
		dfs(i);
		if(flag){
			int cnt=f[0]+f[1]+f[2];
			ans+=1LL*cnt*cnt;
		}else if(f[0]==0||f[1]==0||f[2]==0){
			ans+=f[3];
		}else{
			ans+=1LL*f[0]*f[1];
			ans+=1LL*f[1]*f[2];
			ans+=1LL*f[2]*f[0];
		}
	}
	printf("%lld",ans);
}