#include <bits/stdc++.h>
using namespace std;

int main(){
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n;
	cin >> n;
	vector<int> deg(n, 0);
	for(int i = 0; i < n-1; i++){
		int u, v, a;
		cin >> u >> v >> a;
		deg[u] ^= a;
		deg[v] ^= a;
	}
	int ans = 0;
	int mask = 0;
	for(int x : deg){
		if(x == 0) continue;
		if((mask >> x) & 1) ans += 1;
		mask ^= (1 << x);
	}
	const int B = 4;
	const int F = 1 << B;
	vector<int> dp(1 << F, -1e6);
	vector<bool> ok(1 << F, false);
	for(int i = 0; i < (1 << F); i++){
		int val = 0;
		for(int j = 0; j < F; j++){
			if((i >> j) & 1) val ^= j;
		}
		if(val == 0) ok[i] = true;
	}
	dp[0] = 0;
	for(int j = 0; j < (1 << F); j++){
		int k = j;
		while(k > 0){
			if(ok[k]){
				dp[j] = max(dp[j], dp[j ^ k] + 1);
			}
			k = (k-1) & j;
		}
	}
	cout << (ans + __builtin_popcount(mask) - dp[mask]) << '\n';
}
### Prompt

Your task is to create 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 main(){
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n;
	cin >> n;
	vector<int> deg(n, 0);
	for(int i = 0; i < n-1; i++){
		int u, v, a;
		cin >> u >> v >> a;
		deg[u] ^= a;
		deg[v] ^= a;
	}
	int ans = 0;
	int mask = 0;
	for(int x : deg){
		if(x == 0) continue;
		if((mask >> x) & 1) ans += 1;
		mask ^= (1 << x);
	}
	const int B = 4;
	const int F = 1 << B;
	vector<int> dp(1 << F, -1e6);
	vector<bool> ok(1 << F, false);
	for(int i = 0; i < (1 << F); i++){
		int val = 0;
		for(int j = 0; j < F; j++){
			if((i >> j) & 1) val ^= j;
		}
		if(val == 0) ok[i] = true;
	}
	dp[0] = 0;
	for(int j = 0; j < (1 << F); j++){
		int k = j;
		while(k > 0){
			if(ok[k]){
				dp[j] = max(dp[j], dp[j ^ k] + 1);
			}
			k = (k-1) & j;
		}
	}
	cout << (ans + __builtin_popcount(mask) - dp[mask]) << '\n';
}
```