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 maxn = 1e5 + 100;

int n, m, cnt[3], tot;
vector<int> g[maxn], rg[maxn];
int lvl[maxn];
long long ans;
bool flag;

void dfs(int u, int c) {
	if(~lvl[u]) {
		if(lvl[u] != c) {
			flag = 1;
		}
		return;
	}
	++cnt[lvl[u] = c];
	tot += g[u].size();
	for (int v: g[u]) {
		dfs(v, (c + 1) % 3);
	}
	for (int v: rg[u]) {
		dfs(v, (c + 2) % 3);
	}
	return;
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; ++i) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].push_back(v);
		rg[v].push_back(u);
	}
	memset(lvl + 1, -1, sizeof(lvl[0]) * n);
	for (int i = 1; i <= n; ++i) {
		if(!~lvl[i]) {
			memset(cnt, 0, sizeof(cnt));
			flag = 0;
			tot = 0;
			dfs(i, 0);
			if(flag) {
				cnt[0] += cnt[1] + cnt[2];
				ans += 1ll * cnt[0] * cnt[0];
			}
			else {
				if(cnt[0] && cnt[1] && cnt[2]) {
					ans += 1ll * cnt[0] * (cnt[1] + cnt[2]) + 1ll * cnt[1] * cnt[2];
				}
				else {
					ans += tot;
				}
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}