#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
using namespace std;

typedef long long ll;
typedef double db;
typedef long double ldb;

const int inf=1e9+10;
const int maxn=2e5+10;

int dp[1<<20],n,f[1<<20],vl[maxn],cnt[20];
vector <int> v;

int main(){
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int u,v,c;
		scanf("%d%d%d",&u,&v,&c);
		vl[u]^=c; vl[v]^=c;
	}
	for(int i=0;i<16;i++) cnt[i]=0;
	for(int i=0;i<n;i++){
	//	cout<<vl[i]<<endl; 
		cnt[vl[i]]++;
	}
	int ans=0;
	for(int i=1;i<16;i++){
		ans+=cnt[i]/2;
		if(cnt[i]%2==1){
		//	cout<<i<<endl;
			v.pb(i);
		}
	}
//	cout<<mask<<endl;
	for(int msk=0;msk<(1<<(int)v.size());msk++){
		int val=0;
		for(int i=0;i<v.size();i++){
			if(msk&(1<<i)){
				val^=v[i];
			}
		}
	//	if(val==0) cout<<msk<<endl;
		if(val==0) f[msk]=1;
	}
	for(int msk=0;msk<(1<<(int)v.size());msk++){
		int u=msk;
		while(u>0){
			if(f[u]){
				dp[msk]=max(dp[msk],dp[u^msk]+1);
			}
			u=(u-1)&msk;
		}
	}
	printf("%d\n",ans+(int)v.size()-dp[(1<<(int)v.size())-1]);
	return 0;
}

### Prompt

Please create a solution in Cpp 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<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
using namespace std;

typedef long long ll;
typedef double db;
typedef long double ldb;

const int inf=1e9+10;
const int maxn=2e5+10;

int dp[1<<20],n,f[1<<20],vl[maxn],cnt[20];
vector <int> v;

int main(){
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int u,v,c;
		scanf("%d%d%d",&u,&v,&c);
		vl[u]^=c; vl[v]^=c;
	}
	for(int i=0;i<16;i++) cnt[i]=0;
	for(int i=0;i<n;i++){
	//	cout<<vl[i]<<endl; 
		cnt[vl[i]]++;
	}
	int ans=0;
	for(int i=1;i<16;i++){
		ans+=cnt[i]/2;
		if(cnt[i]%2==1){
		//	cout<<i<<endl;
			v.pb(i);
		}
	}
//	cout<<mask<<endl;
	for(int msk=0;msk<(1<<(int)v.size());msk++){
		int val=0;
		for(int i=0;i<v.size();i++){
			if(msk&(1<<i)){
				val^=v[i];
			}
		}
	//	if(val==0) cout<<msk<<endl;
		if(val==0) f[msk]=1;
	}
	for(int msk=0;msk<(1<<(int)v.size());msk++){
		int u=msk;
		while(u>0){
			if(f[u]){
				dp[msk]=max(dp[msk],dp[u^msk]+1);
			}
			u=(u-1)&msk;
		}
	}
	printf("%d\n",ans+(int)v.size()-dp[(1<<(int)v.size())-1]);
	return 0;
}

```