#include<bits/stdc++.h>
#define vi vector<int>
#define pb push_back
using namespace std;
const int N=1e5+5;
const int M=1<<16+3;
int dp[M],n;
bool good[M];int ans;
int num[N],cnt[20],mask;
void init()
{
	for(int i=1;i<(1<<16);i++)
	{
		int sum=0;
		for(int j=1;j<16;j++)
			if((i>>j)&1)
				sum^=j;
		if(sum==0)
		{
			good[i]=true;
		}
	}
	return;
}//Accepted
 
int main()
{
	cin>>n;
	for(int i=1;i<n;i++)
	{
		int a,b,c;
		cin>>a>>b>>c;
		num[a]^=c;
		num[b]^=c;
	}//Accepted
	for(int i=0;i<n;i++)
		if(num[i])
			cnt[num[i]]++;
	for(int i=1;i<20;i++)
	{
		ans+=cnt[i]/2;
		cnt[i]%=2;
		if(cnt[i])
		{
			ans++;
			mask+=(1<<i);
		}
	}
	init();//Accetped
	for(int i=0;i<=mask;i++)
	{
		for(int j=i;j;j=(j-1)&i)
		{
			if(good[j])
			{
				dp[i]=max(dp[i],dp[i^j]+1);
			}
		}
	}
	cout<<ans-dp[mask]<<endl;
	return 0;
}
### Prompt

Please provide a CPP coded solution 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>
#define vi vector<int>
#define pb push_back
using namespace std;
const int N=1e5+5;
const int M=1<<16+3;
int dp[M],n;
bool good[M];int ans;
int num[N],cnt[20],mask;
void init()
{
	for(int i=1;i<(1<<16);i++)
	{
		int sum=0;
		for(int j=1;j<16;j++)
			if((i>>j)&1)
				sum^=j;
		if(sum==0)
		{
			good[i]=true;
		}
	}
	return;
}//Accepted
 
int main()
{
	cin>>n;
	for(int i=1;i<n;i++)
	{
		int a,b,c;
		cin>>a>>b>>c;
		num[a]^=c;
		num[b]^=c;
	}//Accepted
	for(int i=0;i<n;i++)
		if(num[i])
			cnt[num[i]]++;
	for(int i=1;i<20;i++)
	{
		ans+=cnt[i]/2;
		cnt[i]%=2;
		if(cnt[i])
		{
			ans++;
			mask+=(1<<i);
		}
	}
	init();//Accetped
	for(int i=0;i<=mask;i++)
	{
		for(int j=i;j;j=(j-1)&i)
		{
			if(good[j])
			{
				dp[i]=max(dp[i],dp[i^j]+1);
			}
		}
	}
	cout<<ans-dp[mask]<<endl;
	return 0;
}
```