#include<iostream>
#include<cstdio>
#include<fstream>
#include<cstring>
#define Maxn 100000
using namespace std;
int N;
int dian[Maxn+5]={0};
int cnt[20]={0};
int ans=0;
int f[1<<20];
int Xor[1<<20]={0};
int main()
{
	//memset(f,0x3f,sizeof(f));
	scanf("%d",&N);
	for(int i=1,x,y,c;i<N;i++){
		scanf("%d%d%d",&x,&y,&c);
		x+=1,y+=1;
		dian[x]^=c;
		dian[y]^=c;
	}
	for(int i=1;i<=N;i++)cnt[dian[i]]++;
	int tot=0;
	for(int i=1;i<=15;i++)ans+=(cnt[i]/2),tot|=((1&cnt[i])<<(i-1));
	//for(int i=1;i<=N;i++)cout<<dian[i]<<endl;
	for(int i=1;i<(1<<15);i++)f[i]=f[i>>1]+(i&1);
	for(int i=1;i<(1<<15);i++)f[i]-=1;
	for(int s=1;s<(1<<15);s++){
		for(int i=1;i<=15;i++){
			if((s>>(i-1))&1)
			Xor[s]^=i;
		}
	}
	for(int s=1;s<(1<<15);s++){
		if(Xor[s])continue;
		for(int k=(s-1)&s;k;k=(k-1)&s){
			if(Xor[k])continue;
			f[s]=min(f[s],f[k]+f[s^k]);
		}
	}
	cout<<ans+f[tot];
	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<iostream>
#include<cstdio>
#include<fstream>
#include<cstring>
#define Maxn 100000
using namespace std;
int N;
int dian[Maxn+5]={0};
int cnt[20]={0};
int ans=0;
int f[1<<20];
int Xor[1<<20]={0};
int main()
{
	//memset(f,0x3f,sizeof(f));
	scanf("%d",&N);
	for(int i=1,x,y,c;i<N;i++){
		scanf("%d%d%d",&x,&y,&c);
		x+=1,y+=1;
		dian[x]^=c;
		dian[y]^=c;
	}
	for(int i=1;i<=N;i++)cnt[dian[i]]++;
	int tot=0;
	for(int i=1;i<=15;i++)ans+=(cnt[i]/2),tot|=((1&cnt[i])<<(i-1));
	//for(int i=1;i<=N;i++)cout<<dian[i]<<endl;
	for(int i=1;i<(1<<15);i++)f[i]=f[i>>1]+(i&1);
	for(int i=1;i<(1<<15);i++)f[i]-=1;
	for(int s=1;s<(1<<15);s++){
		for(int i=1;i<=15;i++){
			if((s>>(i-1))&1)
			Xor[s]^=i;
		}
	}
	for(int s=1;s<(1<<15);s++){
		if(Xor[s])continue;
		for(int k=(s-1)&s;k;k=(k-1)&s){
			if(Xor[k])continue;
			f[s]=min(f[s],f[k]+f[s^k]);
		}
	}
	cout<<ans+f[tot];
	return 0;
} 
```