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 <cstdio>

#define iter(i, n) for (int i = 1; i <= n; ++i)
#define iter_e(i, x) for (int i = pos[x]; i; i = g[i].frt)

#define NR 101000
typedef long long i64;

struct Edge { int y, frt; } g[NR * 2];

int n, m, pos[NR], gsz, cnt_e, col[NR];
i64 cnt[3], ans;
bool invalid;

void AE(int x, int y) { g[++gsz] = (Edge) { y, pos[x] }, pos[x] = gsz; }

#define v g[i].y
void dfs(int x) {
	++cnt[col[x]];
	iter_e(i, x) {
		int c = (col[x] + (i % 2 == 1 ? 1 : 2)) % 3;
		if (col[v] == -1) col[v] = c, dfs(v);
		else if (col[v] != c) invalid = true;
		++cnt_e;
	}
}

int main() {
	scanf("%d%d", &n, &m);
	iter(i, m) {
		int x, y;
		scanf("%d%d", &x, &y);
		AE(x, y), AE(y, x);
	}
	iter(i, n) col[i] = -1;
	iter(i, n) if (col[i] == -1) {
		cnt[0] = cnt[1] = cnt[2] = cnt_e = 0; 
		col[i] = 0;
		invalid = false;
		dfs(i);

		//if (cnt[0] + cnt[1] + cnt[2] > 2) printf("%d %lld %lld %lld %d\n", invalid, cnt[0], cnt[1], cnt[2], cnt_e/2);

		if (invalid) ans += (cnt[0] + cnt[1] + cnt[2]) * (cnt[0] + cnt[1] + cnt[2]);
		else if (cnt[1] != 0 && cnt[2] != 0) ans += cnt[0] * cnt[1] + cnt[1] * cnt[2] + cnt[2] * cnt[0];
		else ans += cnt_e / 2;
	}
	printf("%lld\n", ans);
	return 0;
}