#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
using namespace std;
vector<int>bin[20];
const int maxn=1000000;
int n,d[maxn+4]={0},v[20]={0},f[(1<<20)+4];
int main()
{
	int x,y,z;
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&z);
		d[x]^=z,d[y]^=z;
	} 
	int s=0,res=0,t=0;
	for(int i=0;i<n;i++)v[d[i]]++;
	for(int i=1;i<=15;i++){
		res+=v[i]/2,v[i]%=2;
		if(v[i]){
			s|=(1<<(i-1));
			t++;
		}
	}
	memset(f,0x3f,sizeof(f));
	f[0]=0;
	for(int i=0;i<=(1<<15)-1;i++){
		int cnt=0;
		for(int j=0;j<=14;j++)
			if(i&(1<<j))cnt++;
		bin[cnt].push_back(i);
	}
	for(int p=2;p<=t;p++)
		for(int k=0;k<bin[p].size();k++){
			int i=bin[p][k];
			for(int u=0;u<=14;u++)
			if(i&(1<<u))
			for(int v=0;v<=14;v++)
				if(i&(1<<v)&&(u!=v)){
					int w=(u+1)^(v+1)-1; 
					if(u==w||v==w)continue;
					if(i&(1<<w))f[i]=min(f[i],f[i-(1<<w)-(1<<u)-(1<<v)]+2);
					else f[i]=min(f[i],f[i-(1<<u)-(1<<v)+(1<<w)]+1);
				}
		}
	printf("%d",f[s]+res);
	return 0;
}

### Prompt

In CPP, your task is to solve 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<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
using namespace std;
vector<int>bin[20];
const int maxn=1000000;
int n,d[maxn+4]={0},v[20]={0},f[(1<<20)+4];
int main()
{
	int x,y,z;
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&z);
		d[x]^=z,d[y]^=z;
	} 
	int s=0,res=0,t=0;
	for(int i=0;i<n;i++)v[d[i]]++;
	for(int i=1;i<=15;i++){
		res+=v[i]/2,v[i]%=2;
		if(v[i]){
			s|=(1<<(i-1));
			t++;
		}
	}
	memset(f,0x3f,sizeof(f));
	f[0]=0;
	for(int i=0;i<=(1<<15)-1;i++){
		int cnt=0;
		for(int j=0;j<=14;j++)
			if(i&(1<<j))cnt++;
		bin[cnt].push_back(i);
	}
	for(int p=2;p<=t;p++)
		for(int k=0;k<bin[p].size();k++){
			int i=bin[p][k];
			for(int u=0;u<=14;u++)
			if(i&(1<<u))
			for(int v=0;v<=14;v++)
				if(i&(1<<v)&&(u!=v)){
					int w=(u+1)^(v+1)-1; 
					if(u==w||v==w)continue;
					if(i&(1<<w))f[i]=min(f[i],f[i-(1<<w)-(1<<u)-(1<<v)]+2);
					else f[i]=min(f[i],f[i-(1<<u)-(1<<v)+(1<<w)]+1);
				}
		}
	printf("%d",f[s]+res);
	return 0;
}

```