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;
#define MAXN	100005
struct edge {int dest, type; };
int nxt[2][3] = {{1, 2, 0}, {2, 0, 1}};
vector <edge> a[MAXN];
int n, m, type[MAXN];
bool visited[MAXN], illegal;
long long cnt[3], totedges;
void work(int pos) {
	totedges += a[pos].size();
	cnt[type[pos]]++;
	for (unsigned i = 0; i < a[pos].size(); i++)
		if (!visited[a[pos][i].dest]) {
			visited[a[pos][i].dest] = true;
			type[a[pos][i].dest] = nxt[a[pos][i].type][type[pos]];
			work(a[pos][i].dest);
		} else if (type[a[pos][i].dest] != nxt[a[pos][i].type][type[pos]]) illegal = true;
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		a[x].push_back((edge) {y, 0});
		a[y].push_back((edge) {x, 1});
	}
	long long ans = 0;
	for (int i = 1; i <= n; i++) {
		if (visited[i]) continue;
		memset(cnt, 0, sizeof(cnt));
		illegal = false;
		visited[i] = true;
		type[i] = 0;
		totedges = 0;
		work(i);
		if (illegal) ans += (cnt[0] + cnt[1] + cnt[2]) * (cnt[0] + cnt[1] + cnt[2]);
		else {
			if (cnt[0] && cnt[1] && cnt[2]) ans += cnt[0] * cnt[1] + cnt[1] * cnt[2] + cnt[2] * cnt[0];
			else ans += totedges / 2;
		}
	}
	printf("%lld\n", ans);
	return 0;
}