#include<iostream>
#include<cstring>
using namespace std ;
int n ;
const int N = 1e5 + 10 ;
const int M = 1 << 17 ;
int s[N] , S , ans , f[M] , inf , sum[20] ;
 
inline int dfs(int S) {
  if(! S) return 0 ;
  if(f[S] < inf) return f[S] ;
  for(register int i = 0 ; i < 16 ; i ++)
    if(S & (1 << i)) {
      for(register int j = 0 ; j < 16 ; j ++)
        if(i != j && (S & (1 << j))) {
          int p = i ^ j ;
          int x = S ^ (1 << i) ^ (1 << j) ^ (1 << p) ;
          if(S & (1 << p)) { f[S] = min(f[S] , dfs(x) + 2) ; }
          else { f[S] = min(f[S] , dfs(x) + 1) ; }
        }
    }
    return f[S] ;
}
signed main(){
  ios :: sync_with_stdio(false) ;
  cin.tie(0) ;
  cout.tie(0) ;
  cin >> n ;
  for(register int i = 1 ; i < n ; i ++) { int a , b , c ; cin >> a >> b >> c ; s[a] ^= c ; s[b] ^= c ; }
  for(register int i = 0 ; i < n ; i ++) { sum[s[i]] ++ ; }
  for(register int i = 1 ; i < 16 ; i ++)
    ans += (sum[i] >> 1) , S += (1 << i) * (sum[i] & 1) ;
  memset(f , 0x7f , sizeof(f)) ; inf = f[0] ;
  cout << ans + dfs(S) << '\n' ;
}
/*
5
0 1 1
0 2 3
0 3 6
3 4 4
*/
### Prompt

In Cpp, your task is to solve 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<iostream>
#include<cstring>
using namespace std ;
int n ;
const int N = 1e5 + 10 ;
const int M = 1 << 17 ;
int s[N] , S , ans , f[M] , inf , sum[20] ;
 
inline int dfs(int S) {
  if(! S) return 0 ;
  if(f[S] < inf) return f[S] ;
  for(register int i = 0 ; i < 16 ; i ++)
    if(S & (1 << i)) {
      for(register int j = 0 ; j < 16 ; j ++)
        if(i != j && (S & (1 << j))) {
          int p = i ^ j ;
          int x = S ^ (1 << i) ^ (1 << j) ^ (1 << p) ;
          if(S & (1 << p)) { f[S] = min(f[S] , dfs(x) + 2) ; }
          else { f[S] = min(f[S] , dfs(x) + 1) ; }
        }
    }
    return f[S] ;
}
signed main(){
  ios :: sync_with_stdio(false) ;
  cin.tie(0) ;
  cout.tie(0) ;
  cin >> n ;
  for(register int i = 1 ; i < n ; i ++) { int a , b , c ; cin >> a >> b >> c ; s[a] ^= c ; s[b] ^= c ; }
  for(register int i = 0 ; i < n ; i ++) { sum[s[i]] ++ ; }
  for(register int i = 1 ; i < 16 ; i ++)
    ans += (sum[i] >> 1) , S += (1 << i) * (sum[i] & 1) ;
  memset(f , 0x7f , sizeof(f)) ; inf = f[0] ;
  cout << ans + dfs(S) << '\n' ;
}
/*
5
0 1 1
0 2 3
0 3 6
3 4 4
*/
```