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 <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
 
#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)
#define RepG(i, x) for (int i = head[x]; i; i = edge[i].next)
#define v edge[i].to
#define mp(a, b) make_pair(a, b)
 
using namespace std;

typedef long long LL;
const int N = 100010;

struct Edge{ int to, next, c;} edge[N * 2];
int head[N], num;
void add_edge(int a, int b, int c) { edge[++ num] = (Edge){b, head[a], c}, head[a] = num;}

bool vis[N], flag;
int col[N], cnt[4], cn, cm;
void dfs(int x)
{
	vis[x] = true, cnt[col[x]] ++, cn ++;
	RepG(i, x) {
		cm ++;
		int cv = (col[x] + edge[i].c + 3) % 3;
		if (!vis[v]){ col[v] = cv; dfs(v);}
		else if (col[v] != cv) flag = true;
	}
}

int main()
{
	int n, m;
	scanf("%d%d", &n, &m);
	Rep(i, m) {
		int a, b;
		scanf("%d%d", &a, &b);
		add_edge(a, b, 1); add_edge(b, a, -1);
	}
	
	LL ans = 0;
	
	Rep(i, n) if (!vis[i]) {
		cnt[0] = cnt[1] = cnt[2] = cn = cm = 0, flag = false;
		dfs(i);
	//	printf("%d %d %d %d\n", cnt[0], cnt[1], cnt[2], flag);
		if (flag) ans += 1ll * cn * cn;
		else if (!cnt[1] || !cnt[2]) ans += cm / 2;
		else ans += 1ll * cnt[0] * cnt[1] + 1ll * cnt[0] * cnt[2] + 1ll * cnt[1] * cnt[2];
	}
	
	printf("%lld\n", ans);
	
	return 0;
}