#include <bits/stdc++.h>
using namespace std;
const int TOT = 1 << 15, N = 100010;
int dp[TOT],n,val[N],cnt[16],ans,sta,res[TOT];
int main() {
  int a,b,c,tmp;
  scanf("%d",&n);
  for (int i = 1 ; i < n ; ++ i) {
    scanf("%d%d%d",&a,&b,&c);
    a++, b++;
    val[a] ^= c;
    val[b] ^= c;
  }
  for (int i = 1 ; i <= n ; ++ i)
    ++ cnt[val[i]];
  ans = n;
  ans -= cnt[0];
  for (int i = 1 ; i <= 15 ; ++ i)
    ans -= cnt[i]/2, sta |= (cnt[i]&1) << i >> 1;
  for (int i = 0 ; i < TOT ; ++ i) {
    tmp = 0;
    for (int j = 1 ; j <= 15 ; ++ j)
      if ((i >> j-1)&1) tmp ^= j;
    res[i] = tmp;
  }
  for (int i = 1 ; i < TOT ; ++ i) {
    for (int j = i ; j ; j = (j-1) & i)
      if (!res[j]) dp[i] = max(dp[i],dp[i^j] + 1);
  }
  ans -= dp[sta];
  printf("%d\n",ans);
  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;
const int TOT = 1 << 15, N = 100010;
int dp[TOT],n,val[N],cnt[16],ans,sta,res[TOT];
int main() {
  int a,b,c,tmp;
  scanf("%d",&n);
  for (int i = 1 ; i < n ; ++ i) {
    scanf("%d%d%d",&a,&b,&c);
    a++, b++;
    val[a] ^= c;
    val[b] ^= c;
  }
  for (int i = 1 ; i <= n ; ++ i)
    ++ cnt[val[i]];
  ans = n;
  ans -= cnt[0];
  for (int i = 1 ; i <= 15 ; ++ i)
    ans -= cnt[i]/2, sta |= (cnt[i]&1) << i >> 1;
  for (int i = 0 ; i < TOT ; ++ i) {
    tmp = 0;
    for (int j = 1 ; j <= 15 ; ++ j)
      if ((i >> j-1)&1) tmp ^= j;
    res[i] = tmp;
  }
  for (int i = 1 ; i < TOT ; ++ i) {
    for (int j = i ; j ; j = (j-1) & i)
      if (!res[j]) dp[i] = max(dp[i],dp[i^j] + 1);
  }
  ans -= dp[sta];
  printf("%d\n",ans);
  return 0;
}
```