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;

const int maxn = 100010;

typedef long long ll;

ll ans = 0;

int ok = 1, ty = 0, cnt_e = 0;
int cdeg[maxn], rdeg[maxn];
int l[maxn], vis[maxn], col[maxn], cnt[3], e;

int n, m;

struct Edge {
	int v, w, x;
} E[maxn<<1];

inline void addEdge(int u, int v, int w) {
	E[e].v = v; E[e].x = l[u]; E[e].w = w; l[u] = e++;
}

void dfs(int u) {
	if (cdeg[u] && rdeg[u]) ty = 1;
	cnt_e += cdeg[u];
	cnt[col[u]] ++;
	vis[u] = 1;
	for (int p = l[u]; p >= 0; p = E[p].x) {
		int v = E[p].v;
		if (!vis[v]) {
			col[v] = (col[u] + E[p].w + 3) % 3;
			dfs(v);
		} else {
			if (col[v] != (col[u] + E[p].w + 3) % 3) {
				ok = 0;
			}
		}
	}
}

int main() {
	memset(l, -1, sizeof(l));
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		addEdge(u, v, 1);
		addEdge(v, u, -1);
		++ cdeg[u]; ++ rdeg[v];
	}
	for (int i = 1; i <= n; i++) {
		if (!vis[i]) {
			ok = 1; ty = 0; cnt_e = 0;
			cnt[0] = cnt[1] = cnt[2] = 0;
			dfs(i);
			if (!ok) {
				ans += 1LL*(cnt[0]+cnt[1]+cnt[2])*(cnt[0]+cnt[1]+cnt[2]);
			} else {
				if (!ty) {
					ans += cnt_e;
				} else ans += 1LL*cnt[0]*cnt[1]+1LL*cnt[1]*cnt[2]+1LL*cnt[2]*cnt[0];
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}