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 ZeroOneTree {
	public static int[][] father;
	public static int[][] size;
	
	public static int find(int c, int x) {
		 while (father[c][x] != x) {
			 father[c][x] = father[c][father[c][x]];
			 x = father[c][x];
		 }
		 return x;
	}
	
	public static void union(int c, int x, int y) {
		int fcx = find(c, x);
		int fcy = find(c, y);
		if (fcx == fcy) return;
		if (size[c][fcx] > size[c][fcy]) {
			father[c][fcy] = fcx;
			size[c][fcx] += size[c][fcy];
		} else {
			father[c][fcx] = fcy;
			size[c][fcy] += size[c][fcx];
		}
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		father = new int[2][n + 1];
		size = new int[2][n + 1];
		for (int i = 1; i <= n; i++) {
			father[0][i] = i;
			father[1][i] = i;
			size[0][i] = 1;
			size[1][i] = 1;
		}
		for (int i = 0; i < n - 1; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			int c = sc.nextInt();
			union(c, x, y);
		}
		
		long ans = 0;
		for (int i = 1; i <= n; i++) {
			if (father[0][i] == i) ans += 1L * size[0][i] * (size[0][i] - 1);
			if (father[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);
	}
}
