#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n - 1; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    a[x] ^= z;
    a[y] ^= z;
  }
  vector<int> cnt(16);
  for (int i = 0; i < n; i++) {
    ++cnt[a[i]];
  }
  int res = 0;
  int have = 0;
  for (int i = 1; i < 16; i++) {
    res += cnt[i] / 2;
    cnt[i] &= 1;
    have |= (cnt[i] << i);
  }
  vector<bool> ok(1 << 16);
  for (int mask = 0; mask < (1 << 16); mask++) {
    int here = 0;
    for (int i = 0; i < 16; i++) {
      if ((mask >> i) & 1) {
        here ^= i;
      }
    }
    ok[mask] = (here == 0);
  }
  vector<int> dp(1 << 16);
  for (int mask = 1; mask < (1 << 16); mask++) {
    if (mask & 1) continue;
    for (int smsk = mask; smsk > 0; smsk = (smsk - 1) & mask) {
      if (ok[smsk]) {
        dp[mask] = max(dp[mask], dp[mask ^ smsk] + 1);
      }
    }
  }
  cout << res + __builtin_popcount(have) - dp[have] << 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;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n - 1; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    a[x] ^= z;
    a[y] ^= z;
  }
  vector<int> cnt(16);
  for (int i = 0; i < n; i++) {
    ++cnt[a[i]];
  }
  int res = 0;
  int have = 0;
  for (int i = 1; i < 16; i++) {
    res += cnt[i] / 2;
    cnt[i] &= 1;
    have |= (cnt[i] << i);
  }
  vector<bool> ok(1 << 16);
  for (int mask = 0; mask < (1 << 16); mask++) {
    int here = 0;
    for (int i = 0; i < 16; i++) {
      if ((mask >> i) & 1) {
        here ^= i;
      }
    }
    ok[mask] = (here == 0);
  }
  vector<int> dp(1 << 16);
  for (int mask = 1; mask < (1 << 16); mask++) {
    if (mask & 1) continue;
    for (int smsk = mask; smsk > 0; smsk = (smsk - 1) & mask) {
      if (ok[smsk]) {
        dp[mask] = max(dp[mask], dp[mask ^ smsk] + 1);
      }
    }
  }
  cout << res + __builtin_popcount(have) - dp[have] << endl;
  return 0;
}

```