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<bits/stdc++.h>

using namespace std;

int N, M, sz, edges, frq[3], ap[100009];
long long ans = 0;
bool ok = 1;
vector < pair < int, char > > v[100009];

void dfs (int nod, int val)
{
    if (ap[nod] != -1)
    {
        if (ap[nod] != val)
            ok = 0;
        return ;
    }
    ap[nod] = val, frq[val] ++, sz ++;
    for (auto it = v[nod].begin (); it != v[nod].end (); it ++)
        dfs (it->first, (val + it->second) % 3), edges ++;
}

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d %d", &N, &M);
for (int i=1; i<=M; i++)
{
    int x, y;
    scanf ("%d %d", &x, &y);
    v[x].push_back ({y, 1}), v[y].push_back ({x, 2});
}
for (int i=1; i<=N; i++)
    ap[i] = -1;
for (int i=1; i<=N; i++)
    if (ap[i] == -1)
    {
        ok = 1, frq[0] = frq[1] = frq[2] = sz = 0, edges = 0, dfs (i, 0), edges /= 2;
        if (!ok) ans += 1LL * sz * sz;
        else
        {
            if (frq[0] == 0 || frq[1] == 0 || frq[2] == 0) ans += edges;
            else ans += 1LL * frq[0] * frq[1] + 1LL * frq[1] * frq[2] + 1LL * frq[2] * frq[0];
        }
    }
printf ("%lld\n", ans);
return 0;
}
