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;
const int N = 1e5 + 9;
typedef pair<int,int> pii;
vector<pii>V[N];
int n, m, s[3], col[N], c;
long long ans;

bool Dfs (int u) {
	bool o = 0; int v, w;
	++s[col[u]];
	for (auto p : V[u]) {
		v = p.first; w = (p.second + 3 + col[u]) % 3; if (p.second == 1) ++c;
		if (col[v] == -1) col[v] = w, o |= Dfs(v); else if (col[v] != w) o |= 1;
	}
	return o;
}

int main () {
	scanf("%d%d", &n, &m); memset(col, -1, sizeof col);
	for (int i = 1, u, v; i <= m; ++i)
	scanf("%d%d", &u, &v), V[u].emplace_back(pii(v, 1)), V[v].emplace_back(pii(u, -1));
	for (int i = 1; i <= n; ++i) if (col[i] == -1) {
		s[0] = s[1] = s[2] = c = 0;
		if (col[i] = 0, Dfs(i)) ans += 1ll * (s[0] + s[1] + s[2]) * (s[0] + s[1] + s[2]);
		else if (s[0] && s[1] && s[2]) ans += 1ll * s[0] * s[1] + 1ll * s[1] * s[2] + 1ll * s[2] * s[0];
		else ans += c;
	}
	printf("%lld", ans);
	return 0;
}