#include <bits/stdc++.h> 
using namespace std; 
const int maxn = 1e5 + 100; 
const int maxm = 1 << 15 + 10;  

int n, ans, state, tot; 
int val[maxn], cnt[20], f[maxm], s[maxn], sum[maxm]; 
// sum表示每个集合的异或值
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++; 
		val[x] ^= z, val[y] ^= z; 
	}
	for (int i = 1; i <= n; ++i) cnt[val[i]]++; 
	ans = n; ans -= cnt[0]; 
	for (int i = 1; i <= 15; ++i) {
		ans -= cnt[i] / 2 ; 
		state |= (cnt[i] & 1) << (i - 1); 
	}
	tot = 1 << 15; 
	for (int i = 0; i < tot; ++i) {
		int tmp = 0; 	
		for (int j = 1; j <= 15; ++j) {
			if ((i >> (j - 1)) & 1) tmp ^= j; 
		}
		sum[i] = tmp; 
	}
	for (int i = 1; i < tot; ++i) {
		for (int j = i; j; j = (j - 1) & i) {
			if (!sum[i]) f[i] = max(f[i], f[i ^ j] + 1); 
		}
	}
	ans -= f[state]; 
	printf("%d\n", ans); 
	return 0; 
}
### Prompt

Develop a solution in CPP to the problem described below:

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; 
const int maxn = 1e5 + 100; 
const int maxm = 1 << 15 + 10;  

int n, ans, state, tot; 
int val[maxn], cnt[20], f[maxm], s[maxn], sum[maxm]; 
// sum表示每个集合的异或值
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++; 
		val[x] ^= z, val[y] ^= z; 
	}
	for (int i = 1; i <= n; ++i) cnt[val[i]]++; 
	ans = n; ans -= cnt[0]; 
	for (int i = 1; i <= 15; ++i) {
		ans -= cnt[i] / 2 ; 
		state |= (cnt[i] & 1) << (i - 1); 
	}
	tot = 1 << 15; 
	for (int i = 0; i < tot; ++i) {
		int tmp = 0; 	
		for (int j = 1; j <= 15; ++j) {
			if ((i >> (j - 1)) & 1) tmp ^= j; 
		}
		sum[i] = tmp; 
	}
	for (int i = 1; i < tot; ++i) {
		for (int j = i; j; j = (j - 1) & i) {
			if (!sum[i]) f[i] = max(f[i], f[i ^ j] + 1); 
		}
	}
	ans -= f[state]; 
	printf("%d\n", ans); 
	return 0; 
}
```