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, t;
		
		public Edge(int a, int b, int c) {
			u=a;
			v=b;
			t=c;
		}
	}
	
	static HashMap<Integer, ArrayList<Edge>> g;
	static int n;
	static long[] type1;
	static long[] type2;
	static long[] type3;
	static long ans1=0;
	static long ans2=0;
	static long ans3=0;
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		type1 = new long[n];
		type2 = new long[n];
		type3 = new long[n];
		g = new HashMap<>();
		for(int i=0; i<n; i++) g.put(i, new ArrayList<>());
		for(int i=0; i<n-1; i++) {
			int p = in.nextInt()-1;
			int q = in.nextInt()-1;
			int t = in.nextInt();
			Edge e = new Edge(p, q, t);
			g.get(p).add(e);
			g.get(q).add(e);
		}
		
		dfs(0, -1);
		go(0, -1);
		
		System.out.println(ans1+ans2+ans3);
	}
	
	public static void dfs(int u, int p) {
		
		
		for(Edge e: g.get(u)) {
			int v = e.u==u?e.v:e.u;
			if (v==p) continue;
			dfs(v, u);
			if (e.t==0) {
				type1[u] += 1+type1[v];
				type2[u] += 0;
				type3[u] += type2[v]+type3[v];
			} else {
				type1[u] += 0;
				type2[u] += 1+type2[v];
				type3[u] += 0;
			}
		}
	}

	
	public static void go(int u, int p) {
		//now u is the root of the tree
		ans1+=type1[u];
		ans2+=type2[u];
		ans3+=type3[u];
		
		for(Edge e: g.get(u)) {
			int v = e.u==u?e.v:e.u;
			if (v==p) continue;
			
			if (e.t==0) {
				type1[u] -= (1+type1[v]);
				type3[u] -= (type2[v]+type3[v]);
				type1[v] += 1+type1[u];
				type3[v] += type2[u]+type3[u];
				go(v, u);
				type3[v] -= type2[u]+type3[u];
				type1[v] -= (1+type1[u]);
				type3[u] += (type2[v]+type3[v]);
				type1[u] += (1+type1[v]);
			} else {
				type2[u] -= (1+type2[v]);
				type2[v] += (1+type2[u]);
				go(v, u);
				type2[v] -= (1+type2[u]);
				type2[u] += 1+type2[v];
			}
			
			
			
		}
	}
}
