Problem:
Score : 1000 points

Problem Statement

You are given a tree with N vertices. The vertices are numbered 0 through N-1, and the edges are numbered 1 through N-1.
Edge i connects Vertex x_i and y_i, and has a value a_i.
You can perform the following operation any number of times:

Choose a simple path and a non-negative integer x, then for each edge e that belongs to the path, change a_e by executing a_e ← a_e ⊕ x (⊕ denotes XOR).

Your objective is to have a_e = 0 for all edges e.
Find the minimum number of operations required to achieve it.

Constraints

2 ≤ N ≤ 10^5

0 ≤ x_i,y_i ≤ N-1

0 ≤ a_i ≤ 15

The given graph is a tree.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1 a_1
x_2 y_2 a_2
:
x_{N-1} y_{N-1} a_{N-1}

Output

Find the minimum number of operations required to achieve the objective.

Sample Input 1

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

Sample Output 1

3

The objective can be achieved in three operations, as follows:

First, choose the path connecting Vertex 1, 2, and x = 1.

Then, choose the path connecting Vertex 2, 3, and x = 2.

Lastly, choose the path connecting Vertex 0, 4, and x = 4.

Sample Input 2

2
1 0 0

Sample Output 2

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define MAX 100010

using namespace std;

int n;
int n1, n2, n3;
int ans;

bool marc[MAX][4];

vector<int> grafo[MAX][4];

void DFS(int i, int bit, int& szComponent, int& qtdLeafs)
{
	szComponent++;
	if(grafo[i][bit].size() == 1) qtdLeafs++;

	marc[i][bit] = true;

	for(int g = 0 ; g < grafo[i][bit].size() ; g++)
	{
		int prox = grafo[i][bit][g];

		if(marc[prox][bit]) continue;

		DFS(prox , bit , szComponent , qtdLeafs);
	}
}

int main()
{
	scanf("%d",&n);

	for(int g = 0 ; g < n - 1 ; g++)
	{
		scanf("%d %d %d",&n1,&n2,&n3);

		for(int h = 0 ; h < 4 ; h++)
		{
			if( !(n3 & (1 << h)) ) continue;

			grafo[ n1 ][h].push_back( n2 );
			grafo[ n2 ][h].push_back( n1 );
		}
	}

	for(int h = 0 ; h < 4 ; h++)
	{
		for(int g = 1 ; g <= n ; g++)
		{
			if(marc[g][h]) continue;

			int qtdLeafs = 0;
			int szComponent = 0;

			DFS(g , h , szComponent , qtdLeafs);

			if(szComponent > 1) ans += (qtdLeafs + 1)/2;
		}	
	}

	printf("%d\n",ans);
}