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 <bits/stdc++.h>
using namespace std;

const int MAXN = 1e5 + 10;

struct Edge {
	int dest, kind;
	
	Edge() {}
	Edge(int _dest, int _kind) : dest(_dest), kind(_kind) {}
};

int n, m, edge_cnt, cnt, col[MAXN];
vector<Edge> edges[MAXN];
bool vis[MAXN];
long long tag[3];

void dfs(int x) {
	vis[x] = 1;
	++tag[col[x]];
	for (int i = 0; i < edges[x].size(); ++i) {
		int y = edges[x][i].dest, k = edges[x][i].kind;
		if (!vis[y]) {
			col[y] = (col[x] + k) % 3;
			edge_cnt += k == 1;
			dfs(y);
		} else {
			edge_cnt += k == 1;
			if (col[y] != (col[x] + k) % 3) cnt = 1;
		}
	}
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; ++i) {
		int x, y;
		scanf("%d%d", &x, &y);
		edges[x].push_back(Edge(y, 1));
		edges[y].push_back(Edge(x, 2));
	}
	long long ans = 0;
	for (int i = 1; i <= n; ++i)
		if (!vis[i]) {
			edge_cnt = cnt = tag[0] = tag[1] = tag[2] = 0;
			dfs(i);
			if (cnt) ans += (tag[0] + tag[1] + tag[2]) * (tag[0] + tag[1] + tag[2]);
			else {
				if (!tag[0] || !tag[1] || !tag[2]) ans += edge_cnt;
				else ans += tag[0] * tag[1] + tag[1] * tag[2] + tag[0] * tag[2];
			}
		}
	printf("%lld\n", ans);
}