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;

const int DIM = 1e5 + 5;

int col[DIM], cnt[3];
vector<pair<int, int>> edg[DIM];

void dfs(int x, int c, bool &ok, int &nrn, int &nre)
{
    ++cnt[c]; col[x] = c; ++nrn;
    
    for (pair<int, int> y : edg[x]) {
        if (y.second == 1)
            ++nre;
        if (col[y.first] == -1)
            dfs(y.first, (c + y.second + 3) % 3, ok, nrn, nre);
        if (col[y.first] != (c + y.second + 3) % 3)
            ok = false;
    }
    
    return;
}

int main(void)
{
    ios :: sync_with_stdio(false);
    cin.tie(); cout.tie();
    
    int n, m;
    cin >> n >> m;
    
    long long ans = m;
    for (int i = 1; i <= m; ++i) {
        int x, y;
        cin >> x >> y;
        
        edg[x].push_back(make_pair(y,  1));
        edg[y].push_back(make_pair(x, -1));
    }
    
    memset(col + 1, -1, n << 2);
    for (int i = 1; i <= n; ++i) {
        if (col[i] != -1) continue;
        bool ok = true; int nrn = 0, nre = 0;
        
        cnt[0] = cnt[1] = cnt[2] = 0;
        dfs(i, 0, ok, nrn, nre);
        
        if (!ok)
            ans += 1LL * nrn * nrn - nre;
        else
        if (cnt[0] and cnt[1] and cnt[2])
            ans += 1LL * cnt[0] * cnt[1] +
                   1LL * cnt[1] * cnt[2] +
                   1LL * cnt[2] * cnt[0] - nre;
    }
    
    cout << ans << endl;
    return 0;
}