#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define LL long long
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
using namespace std;
int n,a[233333],S,ans,f[66666],sum[20];
int dfs(int S){
	if(!S)return 0;
	if(f[S]<INF)return f[S];
	for(int i=0;i<16;i++)if((S>>i)&1)
		for(int j=0;j<16;j++)if(i!=j&&(S>>j)&1){
			int p=i^j,x=S^(1<<i)^(1<<j)^(1<<p);
			if(S>>p&1)f[S]=min(f[S],dfs(x)+2);
			else f[S]=min(f[S],dfs(x)+1);
		}
	return f[S];
}
int main(){
	cin>>n;
	for(int i=1,x,y,z;i<n;i++){
		cin>>x>>y>>z;
		a[x]^=z,a[y]^=z;
	}
	for(int i=0;i<n;i++)sum[a[i]]++;
	for(int i=1;i<16;i++){
		ans+=(sum[i]>>1);
		S+=(1<<i)*(sum[i]&1);
	}
	memset(f,0x3f,sizeof f);
	ans+=dfs(S);
	cout<<ans<<endl;
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#define pb push_back
#define mp make_pair
#define LL long long
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
using namespace std;
int n,a[233333],S,ans,f[66666],sum[20];
int dfs(int S){
	if(!S)return 0;
	if(f[S]<INF)return f[S];
	for(int i=0;i<16;i++)if((S>>i)&1)
		for(int j=0;j<16;j++)if(i!=j&&(S>>j)&1){
			int p=i^j,x=S^(1<<i)^(1<<j)^(1<<p);
			if(S>>p&1)f[S]=min(f[S],dfs(x)+2);
			else f[S]=min(f[S],dfs(x)+1);
		}
	return f[S];
}
int main(){
	cin>>n;
	for(int i=1,x,y,z;i<n;i++){
		cin>>x>>y>>z;
		a[x]^=z,a[y]^=z;
	}
	for(int i=0;i<n;i++)sum[a[i]]++;
	for(int i=1;i<16;i++){
		ans+=(sum[i]>>1);
		S+=(1<<i)*(sum[i]&1);
	}
	memset(f,0x3f,sizeof f);
	ans+=dfs(S);
	cout<<ans<<endl;
	return 0;
}

```