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 = 100100;
vector<pair<int,int> > e[maxn];
int vis[maxn], label[maxn];
long long cnt[3], tot;
pair<bool, int> dfs(int u, int lab){
	vis[u] = 1;
	label[u] = lab;
	cnt[lab]++;
	tot++;
	bool ok = true;
	int eds = e[u].size();
	for(auto ee : e[u]){
		int v, w; tie(v, w) = ee;
		int nw = (lab + w) % 3;
		if(vis[v]){
			ok = ok && label[v] == nw; 
		} else {
			auto s = dfs(v, nw);
			ok = ok && s.first;
			eds += s.second;
		}
	}
	return make_pair(ok, eds);
}
int main(){
	int n, m; cin >> n >> m;
	for(int i = 0; i < m; i++){
		int u, v; cin >> u >> v;
		e[u].push_back({v, 1});
		e[v].push_back({u, 2});
	}
	long long ans = 0;
	for(int i = 1; i <= n; i++) if(!vis[i]){
		tot = cnt[0] = cnt[1] = cnt[2] = 0;
		auto s = dfs(i, 0);
		if(s.first){
			if(cnt[1] && cnt[2]) ans += cnt[0] * cnt[1] + cnt[1] * cnt[2] + cnt[2] * cnt[0];
			else ans += s.second / 2;
		} else {
			ans += tot * tot;
		}
	}
	cout << ans << endl;
	return 0;
}