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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef pair<int, int>pii;
typedef long long ll;
vector<pii>pat[100000];
int col[100000];
bool flag[100000];
bool isok;
bool cont;
ll cnt[3];
int ec;
void dfs(int node, int c)
{
	if (flag[node])
	{
		if (col[node] != c)cont = true;
		return;
	}
	flag[node] = true;
	col[node] = c;
	cnt[c]++;
	int f1 = 0, f2 = 0;
	for (int i = 0; i < pat[node].size(); i++)
	{
		if (pat[node][i].second == 1)f1++;
		else f2++;
		dfs(pat[node][i].first, (c + pat[node][i].second) % 3);
	}
	ec += pat[node].size();
	if (f1 > 0 && f2 > 0)isok = true;
}
int main()
{
	int num, way;
	scanf("%d%d", &num, &way);
	for (int i = 0; i < way; i++)
	{
		int za, zb;
		scanf("%d%d", &za, &zb);
		za--, zb--;
		pat[za].push_back(make_pair(zb, 1));
		pat[zb].push_back(make_pair(za, 2));
	}
	ll ans = 0;
	for (int i = 0; i < num; i++)
	{
		if (!flag[i])
		{
			isok = cont = false;
			cnt[0] = cnt[1] = cnt[2] = 0;
			ec = 0;
			dfs(i, 0);
			if (!isok)ans += ec / 2;
			else
			{
				if (!cont)ans += cnt[0] * cnt[1] + cnt[1] * cnt[2] + cnt[2] * cnt[0];
				else ans += (cnt[0] + cnt[1] + cnt[2])*(cnt[0] + cnt[1] + cnt[2]);
			}
		}
	}
	printf("%lld\n", ans);
}