#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int cnt[20],v[N];
int dp[1<<20];
int dfs(int s)
{
	if(s==0) return 0;
	if(dp[s]!=-1) return dp[s];
	dp[s]=0x3f3f3f3f;
	for(int i=0;i<16;i++)
	{
		if(s>>i&1)
		{
			for(int j=0;j<16;j++)
			{
				if(s>>j&1)
				{
					int x=j^i;
					int ns=s^(1<<j)^(1<<i)^(1<<x);
					if(__builtin_popcount(ns)<__builtin_popcount(s))
					 dp[s]=min(dp[s],dfs(ns)+1+((s&(1<<x))==(1<<x)));
				}
			}
		}
	}
	return dp[s];
}
int main()
{
	int n;
	cin>>n;
	for(int i=1,x,y,z;i<n;i++)
	{
		cin>>x>>y>>z;
		v[x]^=z;
		v[y]^=z; 
	}
	memset(dp,-1,sizeof dp);
	for(int i=0;i<n;i++)
	 if(v[i]) cnt[v[i]]++;
	int ans=0,st=0;;
	for(int i=1;i<16;i++)
	{
		ans+=cnt[i]/2;
		st|=(cnt[i]&1)<<i;
	}
	ans+=dfs(st);
	cout<<ans<<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>
using namespace std;
const int N=1e5+10;
int cnt[20],v[N];
int dp[1<<20];
int dfs(int s)
{
	if(s==0) return 0;
	if(dp[s]!=-1) return dp[s];
	dp[s]=0x3f3f3f3f;
	for(int i=0;i<16;i++)
	{
		if(s>>i&1)
		{
			for(int j=0;j<16;j++)
			{
				if(s>>j&1)
				{
					int x=j^i;
					int ns=s^(1<<j)^(1<<i)^(1<<x);
					if(__builtin_popcount(ns)<__builtin_popcount(s))
					 dp[s]=min(dp[s],dfs(ns)+1+((s&(1<<x))==(1<<x)));
				}
			}
		}
	}
	return dp[s];
}
int main()
{
	int n;
	cin>>n;
	for(int i=1,x,y,z;i<n;i++)
	{
		cin>>x>>y>>z;
		v[x]^=z;
		v[y]^=z; 
	}
	memset(dp,-1,sizeof dp);
	for(int i=0;i<n;i++)
	 if(v[i]) cnt[v[i]]++;
	int ans=0,st=0;;
	for(int i=1;i<16;i++)
	{
		ans+=cnt[i]/2;
		st|=(cnt[i]&1)<<i;
	}
	ans+=dfs(st);
	cout<<ans<<endl;
	return 0;
}
```