#include<bits/stdc++.h>
using namespace std;
using Int = long long;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;
  
  vector<Int> cnt(n,0);
  for(Int i=1;i<n;i++){
    Int u,v,a;
    cin>>u>>v>>a;
    cnt[u]^=a;
    cnt[v]^=a;
  }
  
  vector<Int> bc(16,0);
  for(Int i=0;i<16;i++) bc[i]=__builtin_popcount(i);
  
  Int ans=n,res=0;
  for(Int i=0;i<n;i++){
    if(!cnt[i]){
      ans--;
      continue;
    }
    if((res>>cnt[i])&1) ans--;
    res^=1<<cnt[i];
  }
  
  vector<Int> dp(1<<16,-1),ok(1<<16,0);
  for(Int i=0;i<(1<<16);i++){
    Int tmp=0;
    for(Int j=0;j<16;j++)
      if((i>>j)&1) tmp^=j;
    ok[i]=tmp==0;
  }
  
  function<Int(Int)> dfs;
  dfs=[&](Int b)->Int{
	if(!b) return 0;
	Int &v=dp[b];
	if(~v) return v;
	v=1;
	for(Int i=b;i>0;i=(i-1)&b){
	  if(i==b) continue;
	  if(ok[i]) chmax(v,dfs(i)+dfs(b^i));
	}
	return v;
      };
  
  cout<<ans-dfs(res)<<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>
using namespace std;
using Int = long long;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;
  
  vector<Int> cnt(n,0);
  for(Int i=1;i<n;i++){
    Int u,v,a;
    cin>>u>>v>>a;
    cnt[u]^=a;
    cnt[v]^=a;
  }
  
  vector<Int> bc(16,0);
  for(Int i=0;i<16;i++) bc[i]=__builtin_popcount(i);
  
  Int ans=n,res=0;
  for(Int i=0;i<n;i++){
    if(!cnt[i]){
      ans--;
      continue;
    }
    if((res>>cnt[i])&1) ans--;
    res^=1<<cnt[i];
  }
  
  vector<Int> dp(1<<16,-1),ok(1<<16,0);
  for(Int i=0;i<(1<<16);i++){
    Int tmp=0;
    for(Int j=0;j<16;j++)
      if((i>>j)&1) tmp^=j;
    ok[i]=tmp==0;
  }
  
  function<Int(Int)> dfs;
  dfs=[&](Int b)->Int{
	if(!b) return 0;
	Int &v=dp[b];
	if(~v) return v;
	v=1;
	for(Int i=b;i>0;i=(i-1)&b){
	  if(i==b) continue;
	  if(ok[i]) chmax(v,dfs(i)+dfs(b^i));
	}
	return v;
      };
  
  cout<<ans-dfs(res)<<endl;
  return 0;
}

```