#include <bits/stdc++.h>
using namespace std;
const int N=100005;
int val[N],w[N],sum[(1<<15)+5],f[(1<<15)+5];
inline int count(int x){
	int cnt=0;while (x) x&=(x-1),cnt++;return cnt;
}
int main (){
	int n;scanf ("%d",&n);
	for (int i=1,u,v,w;i<n;i++){
		scanf ("%d%d%d",&u,&v,&w);
		++u,++v,val[u]^=w,val[v]^=w;
	}
	for (int i=1;i<=n;i++) w[val[i]]++;
	for (int s=0;s<(1<<15);s++)
		for (int i=1;i<=15;i++)
			if (s&(1<<(i-1))) sum[s]^=i;
	int ans=0,to=0;
	for (int s=1;s<=15;s++) ans+=(w[s]>>1),w[s]&=1,to|=w[s]*(1<<(s-1));
	for (int s=1;s<(1<<15);s++){
		if (sum[s]==0) f[s]=count(s)-1;
		else {f[s]=114514;continue;}
		for (int t=s&(s-1);t;t=s&(t-1))
			if (sum[t]==0) f[s]=min(f[s],f[s^t]+f[t]);
	}
	printf ("%d",ans+f[to]);
	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>
using namespace std;
const int N=100005;
int val[N],w[N],sum[(1<<15)+5],f[(1<<15)+5];
inline int count(int x){
	int cnt=0;while (x) x&=(x-1),cnt++;return cnt;
}
int main (){
	int n;scanf ("%d",&n);
	for (int i=1,u,v,w;i<n;i++){
		scanf ("%d%d%d",&u,&v,&w);
		++u,++v,val[u]^=w,val[v]^=w;
	}
	for (int i=1;i<=n;i++) w[val[i]]++;
	for (int s=0;s<(1<<15);s++)
		for (int i=1;i<=15;i++)
			if (s&(1<<(i-1))) sum[s]^=i;
	int ans=0,to=0;
	for (int s=1;s<=15;s++) ans+=(w[s]>>1),w[s]&=1,to|=w[s]*(1<<(s-1));
	for (int s=1;s<(1<<15);s++){
		if (sum[s]==0) f[s]=count(s)-1;
		else {f[s]=114514;continue;}
		for (int t=s&(s-1);t;t=s&(t-1))
			if (sum[t]==0) f[s]=min(f[s],f[s^t]+f[t]);
	}
	printf ("%d",ans+f[to]);
	return 0;
}

```