// Code by H~$~C
#include <bits/stdc++.h>
using namespace std;

static const int Maxn = 100005;

int n;
int a[Maxn];
int dp[Maxn];

int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    ++u, ++v;
    a[u] ^= w, a[v] ^= w;
  }
  
  int ans = 0;
  int allmask = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0) {
      ++ans;
    }
    else if (allmask & (1 << a[i] - 1)) {
      allmask ^= (1 << a[i] - 1);
      ++ans;
    }
    else {
      allmask ^= (1 << a[i] - 1);
    }
  }
  
  dp[0] = 0;
  for (int mask = 1; mask < (1 << 15); ++mask) {
    dp[mask] = -1;
    int xorsum = 0;
    for (int i = 0; i < 15; ++i) {
      if (mask >> i & 1) {
        xorsum ^= (i + 1);
      }
    }
    if (xorsum != 0) {
      continue;
    }
    
    for (int smask = mask; smask; smask = (smask - 1) & mask) {
      if (dp[smask] == -1) continue;
      dp[mask] = max(dp[mask], dp[smask] + dp[mask ^ smask]);
    }
    dp[mask] = max(dp[mask], 1);
  }
  
  printf("%d\n", n - (ans + dp[allmask]));
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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
// Code by H~$~C
#include <bits/stdc++.h>
using namespace std;

static const int Maxn = 100005;

int n;
int a[Maxn];
int dp[Maxn];

int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    ++u, ++v;
    a[u] ^= w, a[v] ^= w;
  }
  
  int ans = 0;
  int allmask = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0) {
      ++ans;
    }
    else if (allmask & (1 << a[i] - 1)) {
      allmask ^= (1 << a[i] - 1);
      ++ans;
    }
    else {
      allmask ^= (1 << a[i] - 1);
    }
  }
  
  dp[0] = 0;
  for (int mask = 1; mask < (1 << 15); ++mask) {
    dp[mask] = -1;
    int xorsum = 0;
    for (int i = 0; i < 15; ++i) {
      if (mask >> i & 1) {
        xorsum ^= (i + 1);
      }
    }
    if (xorsum != 0) {
      continue;
    }
    
    for (int smask = mask; smask; smask = (smask - 1) & mask) {
      if (dp[smask] == -1) continue;
      dp[mask] = max(dp[mask], dp[smask] + dp[mask ^ smask]);
    }
    dp[mask] = max(dp[mask], 1);
  }
  
  printf("%d\n", n - (ans + dp[allmask]));
  return 0;
}

```