#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;

int n,num[100010],tot[100010],ok[100010],dp[100010];

int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<n;i++){
		int a,b,c;
		cin>>a>>b>>c;
		num[a]^=c;num[b]^=c;
	}
	int cnt=0;
	for(int i=1;i<=n;i++){
		if(num[i]==0) cnt++;
		else tot[num[i]]++;
	}
	vector<int> v;
	for(int i=1;i<=15;i++){
		cnt+=tot[i]/2;
		if(tot[i]%2==1){
			v.push_back(i);
		}
	}
	if(v.empty()){
		cout<<n-cnt<<endl;
		return 0;
	}
	int len=v.size();
	for(int i=0;i<(1<<len);i++){
		int sum=0;
		for(int j=1;j<=len;j++) if(i&(1<<(j-1))){
			sum^=v[j-1];
		}
		if(sum==0) ok[i]=1;
	}
	for(int i=0;i<(1<<len)-1;i++){
		int mask=((1<<len)-1)^i;
		for(int j=mask;j>0;j=(j-1)&mask) if(ok[j]){
			dp[i|j]=max(dp[i|j],dp[i]+1);
		}
	}
	cout<<n-cnt-dp[(1<<len)-1]<<endl;
	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
#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;

int n,num[100010],tot[100010],ok[100010],dp[100010];

int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<n;i++){
		int a,b,c;
		cin>>a>>b>>c;
		num[a]^=c;num[b]^=c;
	}
	int cnt=0;
	for(int i=1;i<=n;i++){
		if(num[i]==0) cnt++;
		else tot[num[i]]++;
	}
	vector<int> v;
	for(int i=1;i<=15;i++){
		cnt+=tot[i]/2;
		if(tot[i]%2==1){
			v.push_back(i);
		}
	}
	if(v.empty()){
		cout<<n-cnt<<endl;
		return 0;
	}
	int len=v.size();
	for(int i=0;i<(1<<len);i++){
		int sum=0;
		for(int j=1;j<=len;j++) if(i&(1<<(j-1))){
			sum^=v[j-1];
		}
		if(sum==0) ok[i]=1;
	}
	for(int i=0;i<(1<<len)-1;i++){
		int mask=((1<<len)-1)^i;
		for(int j=mask;j>0;j=(j-1)&mask) if(ok[j]){
			dp[i|j]=max(dp[i|j],dp[i]+1);
		}
	}
	cout<<n-cnt-dp[(1<<len)-1]<<endl;
	return 0;
} 
```