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<int, int> P;
template<typename T> inline void chkmin(T &a, const T &b) { a = a < b ? a : b; }
template<typename T> inline void chkmax(T &a, const T &b) { a = a > b ? a : b; }

const int MAXN = 100005;
int vis[MAXN], cnt[5], clr[MAXN], n, m, flag;
vector<int> A[MAXN], B[MAXN];

void dfs(int u, int c) {
	vis[u] = 1, clr[u] = c, ++cnt[c], ++cnt[3];
	for (int v : A[u]) {
		++cnt[4];
		int d = (c + 1) % 3;
		if (vis[v] && clr[v] != d) flag = 1;
		else if (!vis[v]) dfs(v, d);
	}
	for (int v : B[u]) {
		int d = (c + 2) % 3;
		if (vis[v] && clr[v] != d) flag = 1;
		else if (!vis[v]) dfs(v, d);
	}
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++) {
		int u, v; scanf("%d%d", &u, &v);
		A[u].push_back(v);
		B[v].push_back(u);
	}
	LL res = 0;
	for (int i = 1; i <= n; i++) if (!vis[i]) {
		flag = 0;
		memset(cnt, 0, sizeof(cnt));
		dfs(i, 0);
		int c = 0;
		for (int i = 0; i < 3; i++) c += !cnt[i];
		if (flag) res += (LL)cnt[3] * cnt[3];
		else if (c) res += cnt[4];
		else res += (LL)cnt[0] * cnt[1] + (LL)cnt[1] * cnt[2] + (LL)cnt[0] * cnt[2];
	}
	printf("%lld\n", res);
	return 0;
}