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
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
typedef long long i64;
const int MAXN = 1e5 + 10;
 
i64 answer;
i64 totCnt;
i64 N, M;
i64 check[MAXN];
i64 cnt[3];
 
bool flag;
 
vector<pair<i64, i64> > Adj[MAXN];
 
inline void addEdge(i64 u, i64 v, i64 w)
{
	Adj[u].emplace_back(v, w);
}
 
void DFS(int cur)
{
	cnt[check[cur]] ++;
 
	for(auto x : Adj[cur])
	{
		totCnt += (x.second == 1);
 
		if(check[x.first] != -1)
		{
			if((check[cur] + x.second) % 3 != check[x.first])
				flag = true;
		}
		else
		{
			check[x.first] = (check[cur] + x.second) % 3;
			DFS(x.first);
		}
	}
}
 
int main()
{
	ios::sync_with_stdio(false), cin.tie(nullptr);
 
	cin >> N >> M;
	for (int i = 1; i <= M; i++)
	{
		i64 u, v;
		cin >> u >> v;
		addEdge(u, v, 1);
		addEdge(v, u, 2);
	}
 
	memset(check, -1, sizeof check);
 
	for(int i = 1; i <= N; i++)
	{
		if(check[i] != -1)
			continue;
 
		check[i] = 0, totCnt = 0;
		cnt[0] = cnt[1] = cnt[2] = 0;
		flag = false;
 
		DFS(i);
		if(flag)
			answer += (cnt[0] + cnt[1] + cnt[2]) * (cnt[0] + cnt[1] + cnt[2]);
		else if(!cnt[0] || !cnt[1] || !cnt[2])
			answer += totCnt;
		else
			answer += cnt[0] * cnt[1] + cnt[1] * cnt[2] + cnt[2] * cnt[0];
 
	}
 
	cout << answer << endl ;
 
	return 0;
}