#include<bits/stdc++.h>
using namespace std;
 
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define lg long long
#define db double
#define lb(x) ((x)&-(x))
#define ft first
#define sd second
 
#define HII cerr<<"HI"<<endl
#define LLLINE cerr<<"@@@@@@@@@@@@@@@@@@@@"<<endl
 
template <class _T_>
void read(_T_& d){
	d=0;int f=1;char c=getchar();
	for(;c<'0'||c>'9';c=getchar())if(c=='-')f*=-1;
	for(;c>='0'&&c<='9';c=getchar())d=d*10+c-'0';
	d*=f;
}
 
/************************************************/
#define MN 200005
int n;
int v[MN];
int t[16];
int dp[MN];
int r[MN];
int main(){
	read(n);
	for(int i=1,x,y,z;i<n;++i){
		read(x);read(y);read(z);
		v[x]^=z;v[y]^=z;
	}
	for(int i=0;i<n;++i)++t[v[i]];
	int res=0;
	int h=0;
	for(int i=1;i<16;++i){
		res+=(t[i]+1)/2;t[i]%=2;h|=t[i]*(1<<i);
	}
	memset(dp,-0x3f,sizeof dp);
	dp[0]=0;
	for(int i=0;i<h;++i){
		for(int j=0;j<16;++j)if((1<<j)&i){
			r[i]^=j;
		}
	}
	for(int i=0;i<h;++i)if(dp[i]>=0){
		int S=h-i;
		for(int j=S;j;j=S&(j-1))if(r[j]==0){
			dp[i+j]=max(dp[i]+1,dp[i+j]);
		}
	}
	cout<<res-dp[h];
	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<bits/stdc++.h>
using namespace std;
 
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define lg long long
#define db double
#define lb(x) ((x)&-(x))
#define ft first
#define sd second
 
#define HII cerr<<"HI"<<endl
#define LLLINE cerr<<"@@@@@@@@@@@@@@@@@@@@"<<endl
 
template <class _T_>
void read(_T_& d){
	d=0;int f=1;char c=getchar();
	for(;c<'0'||c>'9';c=getchar())if(c=='-')f*=-1;
	for(;c>='0'&&c<='9';c=getchar())d=d*10+c-'0';
	d*=f;
}
 
/************************************************/
#define MN 200005
int n;
int v[MN];
int t[16];
int dp[MN];
int r[MN];
int main(){
	read(n);
	for(int i=1,x,y,z;i<n;++i){
		read(x);read(y);read(z);
		v[x]^=z;v[y]^=z;
	}
	for(int i=0;i<n;++i)++t[v[i]];
	int res=0;
	int h=0;
	for(int i=1;i<16;++i){
		res+=(t[i]+1)/2;t[i]%=2;h|=t[i]*(1<<i);
	}
	memset(dp,-0x3f,sizeof dp);
	dp[0]=0;
	for(int i=0;i<h;++i){
		for(int j=0;j<16;++j)if((1<<j)&i){
			r[i]^=j;
		}
	}
	for(int i=0;i<h;++i)if(dp[i]>=0){
		int S=h-i;
		for(int j=S;j;j=S&(j-1))if(r[j]==0){
			dp[i+j]=max(dp[i]+1,dp[i+j]);
		}
	}
	cout<<res-dp[h];
	return 0;
}
```