#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<ctime>
#define INF 1e9
using namespace std;
const int maxn=100010;
const double Pi=acos(-1.0);
template<class T>void read(T &x)
{
	x=0;int f=0;char ch=getchar();
	while(ch<'0'||ch>'9') {f|=(ch=='-');ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	x=f?-x:x;
	return;
}
int n,num[maxn],book[21],s;
int f[(1<<17)+10],ans;
int dp(int s)
{
	if(!s)
		return 0;
	if(f[s])
		return f[s];
	f[s]=INF;
	for(int i=1;i<=15;i++)
	{
		if(!(s&(1<<i)))
			continue;
		for(int j=1;j<=15;j++)
		{
			if(i==j||!(s&(1<<j)))
				continue;
			int k=i^j;
			int t=s^(1<<i)^(1<<j)^(1<<k);
			if(s&(1<<k))
				f[s]=min(f[s],dp(t)+2);
			else
				f[s]=min(f[s],dp(t)+1);
		}
	}
	return f[s];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		num[a]^=c;
		num[b]^=c;
	}
	for(int i=0;i<n;i++)
		book[num[i]]++;
	for(int i=1;i<=15;i++)
	{
		ans+=(book[i]>>1);
		s|=((book[i]&1)<<i);//奇数最后剩下
	}
	printf("%d",ans+dp(s));
	return 0;
}
### Prompt

Your task is to create a cpp solution to 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<cstdlib>
#include<cmath>
#include<algorithm>
#include<ctime>
#define INF 1e9
using namespace std;
const int maxn=100010;
const double Pi=acos(-1.0);
template<class T>void read(T &x)
{
	x=0;int f=0;char ch=getchar();
	while(ch<'0'||ch>'9') {f|=(ch=='-');ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	x=f?-x:x;
	return;
}
int n,num[maxn],book[21],s;
int f[(1<<17)+10],ans;
int dp(int s)
{
	if(!s)
		return 0;
	if(f[s])
		return f[s];
	f[s]=INF;
	for(int i=1;i<=15;i++)
	{
		if(!(s&(1<<i)))
			continue;
		for(int j=1;j<=15;j++)
		{
			if(i==j||!(s&(1<<j)))
				continue;
			int k=i^j;
			int t=s^(1<<i)^(1<<j)^(1<<k);
			if(s&(1<<k))
				f[s]=min(f[s],dp(t)+2);
			else
				f[s]=min(f[s],dp(t)+1);
		}
	}
	return f[s];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		num[a]^=c;
		num[b]^=c;
	}
	for(int i=0;i<n;i++)
		book[num[i]]++;
	for(int i=1;i<=15;i++)
	{
		ans+=(book[i]>>1);
		s|=((book[i]&1)<<i);//奇数最后剩下
	}
	printf("%d",ans+dp(s));
	return 0;
}
```