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
// Blackout
// * frank_c1
// * 2017 / 12 / 18
 
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int maxn = (int)(1e5) + 5;
vector<int> nxt[maxn], val[maxn];
int vi[maxn], co[maxn];
int M, S;
LL C[3];

void dfs(int x) {
	vi[x] = 1; ++C[co[x]]; 
	for (int i = 0; i < (int)nxt[x].size(); ++i)
		if (!vi[nxt[x][i]]) {
			co[nxt[x][i]] = (co[x] + val[x][i]) % 3;
			M += (val[x][i] == 1); dfs(nxt[x][i]);
		} else {
			M += (val[x][i] == 1);
			if (co[nxt[x][i]] != (co[x] + val[x][i]) % 3) S = 1;
		}
}

int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; ++i) {
		int u, v;
		scanf("%d%d", &u, &v);
		nxt[u].push_back(v); val[u].push_back(1);
		nxt[v].push_back(u); val[v].push_back(2);
	} LL ret = 0;
	for (int i = 1; i <= n; ++i) if (!vi[i]) {
		M = S = 0; memset(C, 0, sizeof(C)); dfs(i);
		if (S) ret += (C[0] + C[1] + C[2]) * (C[0] + C[1] + C[2]);
		else {
			if (!C[0] || !C[1] || !C[2]) ret += M;
			else ret += C[0] * C[1] + C[1] * C[2] + C[2] * C[0];
		}
	}
	return printf("%lld\n", ret), 0;
}