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 N 100005
using namespace std;
struct edge{
    int t,type,next;
}e[N*2];int ecnt,head[N],du[N];
void addedge(int f,int t){
    e[++ecnt]=(edge){t,1,head[f]};head[f]=ecnt;du[f]++;
    e[++ecnt]=(edge){f,-1,head[t]};head[t]=ecnt;
}
int cor[N],num[6];
void DFS(int u,int col){
    if(cor[u]!=-1){
        if(cor[u]!=col) num[0]=-1e9;
        return;
    }
    cor[u]=col;num[col]++;num[3]++;num[4]+=du[u];
    for(int i=head[u];i;i=e[i].next)
        DFS(e[i].t,(col+e[i].type+3)%3);
}
int n,m;
int main(){
    memset(cor,-1,sizeof(cor));
    scanf("%d%d",&n,&m);
    for(int i=1,f,t;i<=m;i++) scanf("%d%d",&f,&t),addedge(f,t);
    long long ans=0;
    for(int i=1;i<=n;i++)
        if(cor[i]==-1){
            memset(num,0,sizeof(num));
            DFS(i,0);
            if(num[0]<0) ans+=1LL*num[3]*num[3];
            else if(num[0]==0||num[1]==0||num[2]==0) ans+=num[4];
            else ans+=num[0]*1LL*num[1]+num[1]*1LL*num[2]+num[2]*1LL*num[0];
        }
    cout<<ans<<endl;
    return 0;
}