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>
#include<cmath>
using namespace std;
int n,m,x,y;
struct node{int y,nxt,t;
}g[1000010];int h[100010],tot=0;
int cnt[3],col[100010],_point,_link;
bool vis[100010],_is;
void dfs(int x)
{
    cnt[col[x]]++;vis[x]=1;_point++;
    for(int i=h[x];i;i=g[i].nxt)
    {
        _link+=(g[i].t==1);
        if (!vis[g[i].y]) col[g[i].y]=(col[x]+g[i].t+3)%3,dfs(g[i].y);
        else if (col[g[i].y]!=(col[x]+g[i].t+3)%3) _is=0;
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        g[++tot]=node{y,h[x],1};h[x]=tot;
        g[++tot]=node{x,h[y],-1};h[y]=tot;
    }
    long long ans=0;
    for (int i=1;i<=n;i++)
    {
        if (vis[i]) continue;
        _is=1;_point=_link=0;cnt[0]=cnt[1]=cnt[2]=0;
        dfs(i);
        if (!_is) ans+=1ll*_point*_point;
        else if (cnt[0]&&cnt[1]&&cnt[2]) ans+=1ll*cnt[0]*cnt[1]+1ll*cnt[1]*cnt[2]+1ll*cnt[2]*cnt[0];
        else ans+=_link;
    }
    printf("%lld",ans);
    
}