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.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
 
public class P1156D{
    private static class Edge {
        int u, v;
        int w;

        public Edge(int a, int b, int c) {
            u = a;
            v = b;
            w = c;
        }

    }


	private static Scanner in;
	private static int n;
    private static HashMap<Integer, ArrayList<Edge>> g;
    private static long[] type1;
    private static long[] type2;
    private static long[] type3;
    private static long ans = 0;


	public static void main(String[] args) {
		in = new Scanner(System.in);
		
        n = in.nextInt();
        g = new HashMap<>();
        type1 = new long[n];
        type2 = new long[n];
        type3 = new long[n];

        for(int i=0; i<n; i++) {
            g.put(i, new ArrayList<Edge>());
        }

        for(int i=0; i<n-1; i++) {
            int u = in.nextInt()-1;
            int v = in.nextInt()-1;
            int w = in.nextInt();

            g.get(u).add(new Edge(u, v, w));
            g.get(v).add(new Edge(v, u, w));
        }

        dfs(0, -1);

        go(0, -1);

        System.out.println(ans);
		
	}

    public static void dfs(int u, int p) {
        
        for(Edge e: g.get(u)) {
            int v = e.v;
            if (v==p) continue;

            dfs(v, u);

            if (e.w==1) {
                type1[u] += 0;
                type2[u] += 1+type2[v];
                type3[u] += 0; 
            } else {
                type1[u] += 1+type1[v];
                type2[u] += 0;
                type3[u] += type2[v]+type3[v];
            }
        }
    }

    private static void go(int u, int p) {
        //u is at the root
        ans += type1[u]+type2[u]+type3[u];

        for(Edge e: g.get(u)) {
            int v = e.v;
            if (v==p) continue;

            if (e.w==0) {
                type1[u] -= (1+type1[v]);
                type2[u] -= 0;
                type3[u] -= (type2[v]+type3[v]);
                type1[v] += 1+type1[u];
                type2[v] += 0;
                type3[v] += (type2[u]+type3[u]);
                go(v, u);
                type3[v] -= (type2[u]+type3[u]);
                type2[v] -= 0;
                type1[v] -= 1+type1[u];
                type3[u] += (type2[v]+type3[v]);
                type2[u] += 0;
                type1[u] += (1+type1[v]);
            } else {
                type1[u] -= 0;
                type2[u] -= 1+type2[v];
                type3[u] -= 0;
                type1[v] += 0;
                type2[v] += 1+type2[u];
                type3[v] += 0;
                go(v, u);
                type3[v] -= 0;
                type2[v] -= 1+type2[u];
                type1[v] -= 0;
                type3[u] += 0;
                type2[u] += 1+type2[v];
                type1[u] += 0;
            }
        }
    }
 
}
