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
#include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
vector<int>E[101000], L[101000];
int Num[101000], n, m, chk, C[3], cc;
long long Res;
void DFS(int a){
    int i;
    C[Num[a]]++;
    cc += E[a].size();
    for(i=0;i<E[a].size();i++){
        if(Num[E[a][i]] == -1){
            Num[E[a][i]] = (Num[a] + L[a][i])%3;
            DFS(E[a][i]);
        }
        else{
            if(Num[E[a][i]] != (Num[a] + L[a][i])%3)chk = 1;
        }
    }
}
int main(){
    int i, a, b, j;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)Num[i]=-1;
    while(m--){
        scanf("%d%d",&a,&b);
        E[a].push_back(b);L[a].push_back(1);
        E[b].push_back(a);L[b].push_back(2);
    }
    for(i=1;i<=n;i++){
        if(Num[i] == -1){
            Num[i] = 0;
            chk = 0;
            cc = 0;
            for(j=0;j<3;j++)C[j]=0;
            DFS(i);
            if(chk){
                Res += 1ll * (C[0]+C[1]+C[2])*(C[0]+C[1]+C[2]);
            }
            else{
                if(!C[0]||!C[1]||!C[2])Res += cc/2;
                else Res += 1ll*C[0]*(C[1]+C[2])+1ll*C[1]*C[2];
            }
        }
    }
    printf("%lld\n",Res);
}