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>

typedef long long ll;
const int N = 100054, M = 200054;

int V, E, Es = 0;
int to[M], first[N], next[M];
int C[3], CE, col[N];

inline void addedge(int u, int v) {
	to[++Es] = v, next[Es] = first[u], first[u] = Es;
	to[++Es] = u, next[Es] = first[v], first[v] = Es;
}

inline bool dfs(int x) {
	int i, y, c; bool result = true; ++C[col[x]];
	for (i = first[x]; i; i = next[i])
		c = (col[x] + 2 - (i & 1)) % 3, ++CE,
		result &= (~col[y = to[i]] ? col[y] == c : (col[y] = c, dfs(y)));
	return result;
}

int main() {
	int i, u, v; ll ans = 0;
	scanf("%d%d", &V, &E);
	for (i = 0; i < E; ++i) scanf("%d%d", &u, &v), addedge(u, v);
	memset(col, -1, sizeof col);
	for (i = 1; i <= V; ++i) if (!~col[i]) {
		C[0] = C[1] = C[2] = CE = col[i] = 0;
		if (!dfs(i)) ans += ll(C[0] + C[1] + C[2]) * (C[0] + C[1] + C[2]);
		else if (C[0] && C[1] && C[2]) ans += (ll)C[0] * C[1] + ll(C[0] + C[1]) * C[2];
		else ans += CE / 2;
	}
	printf("%lld\n", ans);
	return 0;
}