You are given a tree (an undirected connected acyclic graph) consisting of n vertices and n - 1 edges. A number is written on each edge, each number is either 0 (let's call such edges 0-edges) or 1 (those are 1-edges).

Let's call an ordered pair of vertices (x, y) (x ≠ y) valid if, while traversing the simple path from x to y, we never go through a 0-edge after going through a 1-edge. Your task is to calculate the number of valid pairs in the tree.

Input

The first line contains one integer n (2 ≤ n ≤ 200000) — the number of vertices in the tree.

Then n - 1 lines follow, each denoting an edge of the tree. Each edge is represented by three integers x_i, y_i and c_i (1 ≤ x_i, y_i ≤ n, 0 ≤ c_i ≤ 1, x_i ≠ y_i) — the vertices connected by this edge and the number written on it, respectively.

It is guaranteed that the given edges form a tree.

Output

Print one integer — the number of valid pairs of vertices.

Example

Input


7
2 1 1
3 2 0
4 2 1
5 2 0
6 7 1
7 2 1


Output


34

Note

The picture corresponding to the first example:

<image>
import java.util.*;
import java.io.*;

public class Main {
	static Integer[][] fa = new Integer[2][200001];
	static Integer[][] size = new Integer[2][200001];
	
	public static int find(int c, int x) {
		if (x == fa[c][x]) return x;
		fa[c][x] = find(c, fa[c][x]);
		return fa[c][x];
	}
	
	public static void merge(int c, int x, int y) {
		x = find(c, x);
		y = find(c, y);
		if (x==y) return;
		
		fa[c][x] = y;
		size[c][y] += size[c][x];
	}
	
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		
		int n = in.nextInt();
		for (int i = 1; i <= n; i++) {
			fa[0][i] = fa[1][i] = i;
			size[0][i] = size[1][i] = 1;
		}
		
		for (int i = 1; i < n; i++) {
			int u = in.nextInt(), v = in.nextInt(), c = in.nextInt();
			merge(c, u, v);
		}
		
		long ans = 0;
		for (int i = 1; i <= n; i++) {
			if (fa[0][i] == i) ans += 1L * size[0][i] * (size[0][i] - 1);
			if (fa[1][i] == i) ans += 1L * size[1][i] * (size[1][i] - 1);
			ans += 1L * (size[0][find(0, i)] - 1) * (size[1][find(1, i)] - 1);
		}
		System.out.println(ans);
	}
}