#include<bits/stdc++.h>
using namespace std;
const int M=1000000007;
const int N=200005,E=524288;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
int n,m,i,j,k,u,v,w,f[250005],g[250005],d[250005],num[250005],s,ans,pp[250005],p[250005],t;
bool cmp(int a,int b)
{
	return pp[a]<pp[b];
}
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d %d %d",&u,&v,&w);
		++u,++v;
		d[u]^=w,d[v]^=w;
	}
	for(i=1;i<=n;++i)
		++num[d[i]];
	for(i=0;i<16;++i)
	{
		if(i!=0)
			ans+=num[i]/2;
		num[i]&=1;
		s|=num[i]<<i;
	}
	for(i=0;i<(1<<16);++i)
	{
		p[i]=i;
		for(j=0;j<16;++j)
			if((i>>j)&1)
				++pp[i];
	}
	sort(p,p+(1<<16),cmp);
	for(t=0;t<(1<<16);++t)
	{
		i=p[t];
		if(i==0||i==1)
		{
			f[i]=0;
			continue;
		}
		f[i]=1000000007;
		for(j=1;j<16;++j)
			for(k=1;k<16;++k)
				if(j!=k)
					if(((i>>j)&1)&&((i>>k)&1))
						f[i]=min(f[i],f[i^(1<<j)^(1<<k)^(1<<(j^k))]+1+((i>>(j^k))&1));
	}
	cout<<ans+f[s];
}

### 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;
const int M=1000000007;
const int N=200005,E=524288;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
int n,m,i,j,k,u,v,w,f[250005],g[250005],d[250005],num[250005],s,ans,pp[250005],p[250005],t;
bool cmp(int a,int b)
{
	return pp[a]<pp[b];
}
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d %d %d",&u,&v,&w);
		++u,++v;
		d[u]^=w,d[v]^=w;
	}
	for(i=1;i<=n;++i)
		++num[d[i]];
	for(i=0;i<16;++i)
	{
		if(i!=0)
			ans+=num[i]/2;
		num[i]&=1;
		s|=num[i]<<i;
	}
	for(i=0;i<(1<<16);++i)
	{
		p[i]=i;
		for(j=0;j<16;++j)
			if((i>>j)&1)
				++pp[i];
	}
	sort(p,p+(1<<16),cmp);
	for(t=0;t<(1<<16);++t)
	{
		i=p[t];
		if(i==0||i==1)
		{
			f[i]=0;
			continue;
		}
		f[i]=1000000007;
		for(j=1;j<16;++j)
			for(k=1;k<16;++k)
				if(j!=k)
					if(((i>>j)&1)&&((i>>k)&1))
						f[i]=min(f[i],f[i^(1<<j)^(1<<k)^(1<<(j^k))]+1+((i>>(j^k))&1));
	}
	cout<<ans+f[s];
}

```