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 <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 100005;
const int ESIZE = MAXN << 1;
int n, m;
int et = 1, fir[MAXN], to[ESIZE], ne[ESIZE], len[ESIZE];
int col[MAXN];
long long A[5], ans;

void link(int u, int v) {
    to[++et] = v, ne[et] = fir[u], fir[u] = et;
    to[++et] = u, ne[et] = fir[v], fir[v] = et;
    len[et - 1] = 1, len[et] = -1;
}

bool dfs(int o, int fa, int c) {
    A[col[o] = c]++, A[3]++; bool flag = 1;
    for (int i = fir[o]; i; i = ne[i]) {
        A[4]++;
        if (i != fa) {
            if (col[to[i]] == -1)
                flag &= dfs(to[i], i ^ 1, (c + len[i] + 3) % 3);
            else if (col[to[i]] != (c + len[i] + 3) % 3)
                flag = 0;
        }
    }
    return flag;
}

int main() {
    scanf("%d %d", &n, &m);
	for (int i = 1, x, y; i <= m; i++) {
		scanf("%d %d", &x, &y);
        link(x, y);
    }
    for (int i = 1; i <= n; i++)
        col[i] = -1;
    for (int i = 1, res; i <= n; i++)
        if (col[i] == -1) {
            A[0] = A[1] = A[2] = A[3] = A[4] = 0;
            res = dfs(i, 0, 0);
            if (res)
                ans += A[0] && A[1] && A[2] ? A[0] * A[1] + A[1] * A[2] + A[2] * A[0] : A[4] >> 1;
            else
                ans += A[3] * A[3];
        }
    printf("%lld", ans);
    return 0;
}