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>
using namespace std;
typedef long long ll;
const int N=1e5+10;
int n,m,tp[N],c[3],v,e;
vector<int> I[N],O[N];
bool ok,vis[N];
ll ans;
void dfs(int x,int h){
    tp[x]=h%3;
    c[tp[x]]++;
    vis[x]=1;
    v++;
    e+=O[x].size();
    for (int i=0;i<O[x].size();i++){
        int v=O[x][i];
        if (!vis[v]) dfs(v,h+1);else
        if ((tp[v]-tp[x]+3)%3!=1) ok=0;
    }
    for (int i=0;i<I[x].size();i++){
        int v=I[x][i];
        if (!vis[v]) dfs(v,h+2);else
        if ((tp[v]-tp[x]+3)%3!=2) ok=0;
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        O[u].push_back(v);
        I[v].push_back(u);
    }
    for (int i=1;i<=n;i++)
    if (!vis[i]){
        ok=1;c[0]=c[1]=c[2]=v=e=0;
        dfs(i,0);
        if (!ok) ans+=1ll*v*v;else
        if (c[0]&&c[1]&&c[2]) ans+=1ll*c[0]*c[1]+1ll*c[1]*c[2]+1ll*c[2]*c[0];
        else ans+=e;
    }
    printf("%lld\n",ans);
    return 0;
}