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: Accepted)

#include <cstdio>
#include <algorithm>

using namespace std;

const int NMAX = 1e5;
int n;
int b[NMAX + 5];
int dp[(1 << 15)];

int main(){
	
	scanf("%d",&n);
	
	for(int i = 1;i < n;i++){
		int x,y,z;
		scanf("%d %d %d",&x,&y,&z);
		x++;
		y++;
		b[x] ^= z;
		b[y] ^= z;
	}
	
	int ans = n;
	int mask = 0;
	
	
	for(int i = 1;i <= n;i++){
		if(b[i] == 0){
			ans--;
		}
		else if((mask >> (b[i] - 1)) & 1){
			mask ^= (1 << (b[i] - 1));
			ans--;
		}
		else{
			mask ^= (1 << (b[i] - 1));
		}
	}
		
	dp[0] = 0;
	
	for(int i = 1;i <= mask;i++){
		dp[i] = -(1 << 28);
		for(int j = i;j;j = (j - 1) & i){
			int xo = 0;
			
			for(int b = 0;b < 15;b++){
				if((j >> b) & 1){
					xo ^= (b + 1);///cause i subtracted 1 before to make them in [0,15) range
				}
			}
			
			if(xo == 0){
				dp[i] = max(dp[i],1 + dp[i ^ j]);
			}
		}
	}
	
	ans -= dp[mask];
	
	printf("%d",ans);
	
	return 0;
}