#include <bits/stdc++.h>
using namespace std;
 
#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")
 
#define F first
#define S second
#define pb push_back
#define SZ(x) (ll)(x.size())
#define all(x) x.begin(),x.end()
#define MP make_pair
 
typedef int ll;
typedef long double ld;
typedef pair<ll,ll> pll;
 
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll maxn=1e5+10, maxm=5e6+10, lg=16, mod=998244353, inf=1e18;

ll n,tt[lg],x[maxn],ans,dp[1<<lg],xo[1<<lg];

int main(){
	ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	
	cin>>n;
	for(int i=1;i<n;i++){
		ll v,u,a; cin>>v>>u>>a;
		x[v]^=a, x[u]^=a;
	}
	for(int i=0;i<n;i++) tt[x[i]]++;
	vector<ll> a;
	ans+=tt[0];
	for(int i=1;i<lg;i++){
		ans+=tt[i]/2;
		if(tt[i]%2) a.pb(i);
	}
	for(int msk=1;msk<(1<<SZ(a));msk++){
		for(int i=0;i<SZ(a);i++)if(msk>>i&1){ xo[msk]=xo[msk^(1<<i)]^a[i]; break; }
		for(int sub=msk;sub;sub=(sub-1)&msk)if(xo[sub]==0) dp[msk]=max(dp[msk],1+dp[msk^sub]);
	}
	cout<<n-(ans+dp[(1<<SZ(a))-1]);
	
	return 0;
}




### Prompt

Develop a solution in CPP to the problem described below:

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;
 
#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")
 
#define F first
#define S second
#define pb push_back
#define SZ(x) (ll)(x.size())
#define all(x) x.begin(),x.end()
#define MP make_pair
 
typedef int ll;
typedef long double ld;
typedef pair<ll,ll> pll;
 
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll maxn=1e5+10, maxm=5e6+10, lg=16, mod=998244353, inf=1e18;

ll n,tt[lg],x[maxn],ans,dp[1<<lg],xo[1<<lg];

int main(){
	ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	
	cin>>n;
	for(int i=1;i<n;i++){
		ll v,u,a; cin>>v>>u>>a;
		x[v]^=a, x[u]^=a;
	}
	for(int i=0;i<n;i++) tt[x[i]]++;
	vector<ll> a;
	ans+=tt[0];
	for(int i=1;i<lg;i++){
		ans+=tt[i]/2;
		if(tt[i]%2) a.pb(i);
	}
	for(int msk=1;msk<(1<<SZ(a));msk++){
		for(int i=0;i<SZ(a);i++)if(msk>>i&1){ xo[msk]=xo[msk^(1<<i)]^a[i]; break; }
		for(int sub=msk;sub;sub=(sub-1)&msk)if(xo[sub]==0) dp[msk]=max(dp[msk],1+dp[msk^sub]);
	}
	cout<<n-(ans+dp[(1<<SZ(a))-1]);
	
	return 0;
}




```