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 maxn 100009
#define LL long long
using namespace std;
struct data{
  int nex,to,w;
}e[maxn*2];
int head[maxn],edge=0,tot,t[3],co[maxn],flag;
inline void add(int from,int to,int w){
  e[++edge]=(data){head[from],to,w};
  head[from]=edge;
}
void dfs(int x){
  t[co[x]]++;
  for(int i=head[x];i;i=e[i].nex){
    int v=e[i].to;if(e[i].w==1)tot++;
    if(co[v]==-1) co[v]=(co[x]+e[i].w+3)%3,dfs(v);
    else if(co[v]!=(co[x]+e[i].w+3)%3) flag=1;
  }
}
int main(){
  int n,m;LL ans=0;
  scanf("%d%d",&n,&m);
  for(int i=1,x,y;i<=m;i++)
    scanf("%d%d",&x,&y),add(x,y,1),add(y,x,-1);
  memset(co,-1,sizeof(co));
  for(int i=1;i<=n;i++)
    if(co[i]==-1){
      t[0]=t[1]=t[2]=tot=flag=0;
      co[i]=0;dfs(i);
      if(flag) ans+=(LL)(t[1]+t[2]+t[0])*(t[1]+t[2]+t[0]);
      else if(t[0] && t[1] && t[2]) ans+=(LL)t[0]*t[1]+(LL)t[1]*t[2]+(LL)t[2]*t[0];
      else ans+=tot;
    }
  printf("%lld",ans);
  return 0;
}