#include <cstdio>
using namespace std ;
const int MAXN = 1e5 + 10 ;
int n , a[MAXN] , cnt[MAXN] , sum[1 << 16] , f[1 << 16] , ans , t ;
int min (int a , int b) {
	return a < b ? a : b ;
}
int main () {
	scanf ("%d" , &n) ;
	for (int i = 1 ; i < n ; i++) {
		int u , v , w ;
		scanf ("%d %d %d" , &u , &v , &w) ;
		a[u] ^= w ; a[v] ^= w ;
	}
	for (int i = 0 ; i < n ; i++) cnt[a[i]]++ ;
	for (int i = 1 ; i < 16 ; i++) {
		ans += (cnt[i] >> 1) ;
		if (cnt[i] & 1) t |= (1 << (i - 1)) ;
	}
	for (int i = 1 ; i < (1 << 15) ; i++)
		f[i] = f[i >> 1] + (i & 1) ;
	for (int i = 1 ; i < (1 << 15) ; i++) f[i]-- ;
	for (int i = 1 ; i < (1 << 15) ; i++)
		for (int j = 0 ; j < 15 ; j++)
			if (i & (1 << j)) 
				sum[i] ^= (j + 1) ;
	for (int i = 1 ; i < (1 << 15) ; i++)
		if (!sum[i])
			for (int j = (i - 1) & i ; j ; j = (j - 1) & i)
				if (!sum[j])
					f[i] = min (f[i] , f[j] + f[i ^ j]) ;
	printf ("%d\n" , ans + f[t]) ;
	return 0 ;
}
### Prompt

In CPP, your task is to solve 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 <cstdio>
using namespace std ;
const int MAXN = 1e5 + 10 ;
int n , a[MAXN] , cnt[MAXN] , sum[1 << 16] , f[1 << 16] , ans , t ;
int min (int a , int b) {
	return a < b ? a : b ;
}
int main () {
	scanf ("%d" , &n) ;
	for (int i = 1 ; i < n ; i++) {
		int u , v , w ;
		scanf ("%d %d %d" , &u , &v , &w) ;
		a[u] ^= w ; a[v] ^= w ;
	}
	for (int i = 0 ; i < n ; i++) cnt[a[i]]++ ;
	for (int i = 1 ; i < 16 ; i++) {
		ans += (cnt[i] >> 1) ;
		if (cnt[i] & 1) t |= (1 << (i - 1)) ;
	}
	for (int i = 1 ; i < (1 << 15) ; i++)
		f[i] = f[i >> 1] + (i & 1) ;
	for (int i = 1 ; i < (1 << 15) ; i++) f[i]-- ;
	for (int i = 1 ; i < (1 << 15) ; i++)
		for (int j = 0 ; j < 15 ; j++)
			if (i & (1 << j)) 
				sum[i] ^= (j + 1) ;
	for (int i = 1 ; i < (1 << 15) ; i++)
		if (!sum[i])
			for (int j = (i - 1) & i ; j ; j = (j - 1) & i)
				if (!sum[j])
					f[i] = min (f[i] , f[j] + f[i ^ j]) ;
	printf ("%d\n" , ans + f[t]) ;
	return 0 ;
}
```