/*********************
*  Author: xuziyuan  * 
*********************/
 
#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair
 
using namespace std;
 
const LL MOD=1e9+7;

int n,x,y,z,val,num[100010],all,good[70000],dp[70000],ans=0;
vector <int> g[100010];

int main()
{
	cin>>n;
	rep(i,n-1)
	{
		cin>>x>>y>>z;
		g[x].pb(z);
		g[y].pb(z);
	}
	rep(i,n)
	{
		val=0;
		rep(j,g[i].size()) val^=g[i][j];
		num[val]++;
	}
	repn(i,15)
	{
		ans+=num[i]/2;
		if(num[i]%2==1) all|=(1<<i);
	}
	rep(i,1<<16)
	{
		int tmp=0;
		rep(j,16) if((i&(1<<j))>0) tmp^=j;
		if(tmp==0) good[i]=true;
	}
	rep(i,1<<16)
	{
		if(!good[i]) continue;
		for(int j=i;j>0;j=(j-1)&i) if(good[j]) dp[i]=max(dp[i],dp[i^j]+1);
	}
	cout<<ans+__builtin_popcount(all)-dp[all]<<endl;
	return 0;
}
### Prompt

Please formulate 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
/*********************
*  Author: xuziyuan  * 
*********************/
 
#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair
 
using namespace std;
 
const LL MOD=1e9+7;

int n,x,y,z,val,num[100010],all,good[70000],dp[70000],ans=0;
vector <int> g[100010];

int main()
{
	cin>>n;
	rep(i,n-1)
	{
		cin>>x>>y>>z;
		g[x].pb(z);
		g[y].pb(z);
	}
	rep(i,n)
	{
		val=0;
		rep(j,g[i].size()) val^=g[i][j];
		num[val]++;
	}
	repn(i,15)
	{
		ans+=num[i]/2;
		if(num[i]%2==1) all|=(1<<i);
	}
	rep(i,1<<16)
	{
		int tmp=0;
		rep(j,16) if((i&(1<<j))>0) tmp^=j;
		if(tmp==0) good[i]=true;
	}
	rep(i,1<<16)
	{
		if(!good[i]) continue;
		for(int j=i;j>0;j=(j-1)&i) if(good[j]) dp[i]=max(dp[i],dp[i^j]+1);
	}
	cout<<ans+__builtin_popcount(all)-dp[all]<<endl;
	return 0;
}
```