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;
#define LL long long
int n, m;
const int N = 200000;
vector <int> bi[N], bb[N];
int col[N], cnt[3], all;
LL ans; int nw;
void dfs(int t, int c)
{
    if (col[t] == -1)
    {
        col[t] = c;
        cnt[c] ++;
    }
    else
    {
        if (col[t] != c) all = 1;
        return;
    }
    for (int i = 0; i < bi[t].size(); ++ i)
        dfs(bi[t][i], (c + 1) % 3), nw ++;
    for (int i = 0; i < bb[t].size(); ++ i)
        dfs(bb[t][i], (c + 2) % 3);
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; ++ i)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        bi[a].push_back(b);
        bb[b].push_back(a);
    }
    for (int i = 1; i <= n; ++ i) col[i] = -1;
    for (int i = 1; i <= n; ++ i) if (col[i] == -1)
    {
        cnt[0] = cnt[1] = cnt[2] = 0;
        all = nw = 0;
        dfs(i, 0);
        if (all) ans += 1ll * (cnt[0] + cnt[1] + cnt[2]) * (cnt[0] + cnt[1] + cnt[2]);
        else if (cnt[0] && cnt[1] && cnt[2])
            ans += 1ll * cnt[0] * cnt[1] + 1ll * cnt[0] * cnt[2] + 1ll * cnt[1] * cnt[2];
        else ans += nw;
    }
    printf("%lld\n", ans);
}