#include <bits/stdc++.h>

using namespace std;

int D[101010], P[101010];
int T[101010], K[16];
int n, k, ans, s;

int main()
{
	int i, j, t, a, b, c;
	
	scanf("%d", &n);
	
	for(i=1; i<n; i++){
		scanf("%d%d%d", &a, &b, &c);
		P[a] ^= c;
		P[b] ^= c;
	}
	
	for(i=0; i<n; i++){
		K[P[i]] ++;
	}
	
	for(i=0; i<15; i++){
		ans += K[i + 1] >> 1;
		k |= (1 << i) * (K[i + 1] & 1);
		s += (K[i + 1] & 1);
	}
	
	D[0] = 0;
	
	for(i=1; i<(1<<15); i++){
		for(j=t=0; j<15; j++){
			if(i & (1 << j)) t ^= (j + 1);
		}
		if(t == 0) T[i] = 1;
		
		D[i] = 0;
		
		for(j=i; j; j=j-1&i){
			if(T[j]) D[i] = max(D[i], D[i - j] + 1);
		}
	}
	
	printf("%d\n", ans + s - D[k]);
	
	return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

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.

Examples

Input

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


Output

3


Input

2
1 0 0


Output

0

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

int D[101010], P[101010];
int T[101010], K[16];
int n, k, ans, s;

int main()
{
	int i, j, t, a, b, c;
	
	scanf("%d", &n);
	
	for(i=1; i<n; i++){
		scanf("%d%d%d", &a, &b, &c);
		P[a] ^= c;
		P[b] ^= c;
	}
	
	for(i=0; i<n; i++){
		K[P[i]] ++;
	}
	
	for(i=0; i<15; i++){
		ans += K[i + 1] >> 1;
		k |= (1 << i) * (K[i + 1] & 1);
		s += (K[i + 1] & 1);
	}
	
	D[0] = 0;
	
	for(i=1; i<(1<<15); i++){
		for(j=t=0; j<15; j++){
			if(i & (1 << j)) t ^= (j + 1);
		}
		if(t == 0) T[i] = 1;
		
		D[i] = 0;
		
		for(j=i; j; j=j-1&i){
			if(T[j]) D[i] = max(D[i], D[i - j] + 1);
		}
	}
	
	printf("%d\n", ans + s - D[k]);
	
	return 0;
}
```