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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;

#define A first
#define B second

const ll MAXN = 1e5+5;

ll n, ans, m;
vector<pii> g[MAXN];

ll ct[5];
bool y = 0;
ll vis[MAXN];
ll edges = 0, vtx = 0;

void dfs(ll v, ll col) {
	ct[col]++;
	vis[v] = col;
	edges += g[v].size();
	vtx++;

	for (pii t : g[v]) {
		ll x = t.B;
		ll colnext = col;
		if (x == -1) {
			colnext--;
			if (colnext == 0)
				colnext = 3;
		} else {
			colnext++;
			if (colnext == 4)
				colnext = 1;
		}

		if (vis[t.A] != 0) {
			if (vis[t.A] != colnext)
				y = 1;
		} else
			dfs(t.A, colnext);
	}
}

int main() {
	cin >> n >> m;
	for (ll i = 0; i < m; ++i) {
		ll a, b;
		cin >> a >> b;
		a--; b--;

		g[a].push_back(pii(b, 1));
		g[b].push_back(pii(a, -1));
	}

	for (ll i = 0; i < n; ++i) {
		if (vis[i] != 0)
			continue;
		edges = vtx = y = 0;
		ct[1] = ct[2] = ct[3] = 0;
		dfs(i, 1);
		edges /= 2;

		if (y == 1) {
			ans += vtx*vtx;
		} else {
			if (ct[1] == 0 || ct[2] == 0 || ct[3] == 0)
				ans += edges;
			else
				ans += (ct[1] * ct[2] + ct[2] * ct[3] + ct[3] * ct[1]);
		}
	}

	cout << ans << endl;
}