#include<bits/stdc++.h>
using namespace std;
int a[100010],t[16],ans,S,f[1100010];
vector<int>g[16];
int main()
{
	memset(f,0x3f,sizeof(f));
	int n;scanf("%d",&n);
	for(int i=1,u,v,w;i<n;i++)scanf("%d%d%d",&u,&v,&w),a[u]^=w,a[v]^=w;
	for(int i=0;i<n;i++)t[a[i]]++;
	for(int i=1;i<16;i++)ans+=(t[i]>>1),S|=(t[i]&1)<<(i-1);
	for(int i=0;i<=S;i++)
	{
		int x=i,ss=0;
		while(x)
		{
			if(x&1)ss++; 
			x>>=1;
		}
		g[ss].push_back(i);
	} 
	f[S]=0;
	for(int p=15;p>=2;p--)
		for(int e=0,s;e<g[p].size();e++)
		{
			s=g[p][e];if(f[s]==f[S+1])continue;
			for(int i=1;i<16;i++)if(s&(1<<(i-1)))for(int j=1;j<16;j++)if(i!=j&&(s&(1<<(j-1))))
			{
				int a=i^j; 
				if(s&(1<<(a-1)))f[s-(1<<(i-1))-(1<<(j-1))-(1<<(a-1))]=min(f[s-(1<<(i-1))-(1<<(j-1))-(1<<(a-1))],f[s]+2);
				else f[s-(1<<(i-1))-(1<<(j-1))+(1<<(a-1))]=min(f[s-(1<<(i-1))-(1<<(j-1))+(1<<(a-1))],f[s]+1);
			}
		}
	printf("%d",f[0]+ans);
	return 0;
}
### Prompt

Create a solution in Cpp for 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 a[100010],t[16],ans,S,f[1100010];
vector<int>g[16];
int main()
{
	memset(f,0x3f,sizeof(f));
	int n;scanf("%d",&n);
	for(int i=1,u,v,w;i<n;i++)scanf("%d%d%d",&u,&v,&w),a[u]^=w,a[v]^=w;
	for(int i=0;i<n;i++)t[a[i]]++;
	for(int i=1;i<16;i++)ans+=(t[i]>>1),S|=(t[i]&1)<<(i-1);
	for(int i=0;i<=S;i++)
	{
		int x=i,ss=0;
		while(x)
		{
			if(x&1)ss++; 
			x>>=1;
		}
		g[ss].push_back(i);
	} 
	f[S]=0;
	for(int p=15;p>=2;p--)
		for(int e=0,s;e<g[p].size();e++)
		{
			s=g[p][e];if(f[s]==f[S+1])continue;
			for(int i=1;i<16;i++)if(s&(1<<(i-1)))for(int j=1;j<16;j++)if(i!=j&&(s&(1<<(j-1))))
			{
				int a=i^j; 
				if(s&(1<<(a-1)))f[s-(1<<(i-1))-(1<<(j-1))-(1<<(a-1))]=min(f[s-(1<<(i-1))-(1<<(j-1))-(1<<(a-1))],f[s]+2);
				else f[s-(1<<(i-1))-(1<<(j-1))+(1<<(a-1))]=min(f[s-(1<<(i-1))-(1<<(j-1))+(1<<(a-1))],f[s]+1);
			}
		}
	printf("%d",f[0]+ans);
	return 0;
}
```