#include <bits/stdc++.h>
#define INF 1000000000
#define LINF 1000000000000000000
#define MOD 1000000007
#define mod 1000007
#define INF63 1061109567
#define INF127 9187201950435737471
#define F first
#define S second
#define ll long long
#define N 100010
#define M 20
#define NM 70000
using namespace std;
int n,m,val[N],tms[M],v=0,ans=0,dp[NM];
bool can[NM]={};
int main(){
	int i,j,x,y,z;
	cin>>n;
	memset(tms,0,sizeof(tms));
	memset(val,0,sizeof(val));
	for(i=1;i<n;i++)
	{
		cin>>x>>y>>z;
		val[x]^=z;
		val[y]^=z;
	}
	for(i=0;i<n;i++)
	{
		tms[val[i]]++;
	}
	ans+=tms[0];
	for(i=1;i<16;i++)
	{
		if(tms[i]%2==1)
		{
			v+=(1<<i);
		}
		ans+=tms[i]/2;
	}
	for(i=0;i<(1<<16);i++)
	{
		x=0;
		for(j=1;j<16;j++)
		{
			if((i&(1<<j))!=0)
			{
				x^=j;
			}
		}
		if(x==0)
		{
			can[i]=true;
		}
	}
	memset(dp,0,sizeof(dp));
	for(i=0;i<(1<<16);i++)
	{
		for(j=i;j!=0;j=i&(j-1))
		{
			if(can[j])
			{
				dp[i]=max(dp[i],dp[j]+1);
			}
		}
	}
	ans+=dp[v];
	cout<<n-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 INF 1000000000
#define LINF 1000000000000000000
#define MOD 1000000007
#define mod 1000007
#define INF63 1061109567
#define INF127 9187201950435737471
#define F first
#define S second
#define ll long long
#define N 100010
#define M 20
#define NM 70000
using namespace std;
int n,m,val[N],tms[M],v=0,ans=0,dp[NM];
bool can[NM]={};
int main(){
	int i,j,x,y,z;
	cin>>n;
	memset(tms,0,sizeof(tms));
	memset(val,0,sizeof(val));
	for(i=1;i<n;i++)
	{
		cin>>x>>y>>z;
		val[x]^=z;
		val[y]^=z;
	}
	for(i=0;i<n;i++)
	{
		tms[val[i]]++;
	}
	ans+=tms[0];
	for(i=1;i<16;i++)
	{
		if(tms[i]%2==1)
		{
			v+=(1<<i);
		}
		ans+=tms[i]/2;
	}
	for(i=0;i<(1<<16);i++)
	{
		x=0;
		for(j=1;j<16;j++)
		{
			if((i&(1<<j))!=0)
			{
				x^=j;
			}
		}
		if(x==0)
		{
			can[i]=true;
		}
	}
	memset(dp,0,sizeof(dp));
	for(i=0;i<(1<<16);i++)
	{
		for(j=i;j!=0;j=i&(j-1))
		{
			if(can[j])
			{
				dp[i]=max(dp[i],dp[j]+1);
			}
		}
	}
	ans+=dp[v];
	cout<<n-ans<<endl;
	return 0;
}
```