#include <bits/stdc++.h>

using namespace std;

typedef vector<int> Vec;

vector<int> gay;

int Dfs(int msk) {
  if (!msk) return 0;
  assert((msk & -msk) != msk);
  int &res = gay[msk];
  if (res) return res;
  res = 2233;
  for (int i = 0; i < 16; ++i) {
    if (msk >> i & 1) {
      for (int j = i + 1; j < 16; ++j) {
        if (msk >> j & 1) {
          int g = i ^ j;
          res = min(res, 1 + (msk >> g & 1) + Dfs(msk ^ (1 << i) ^ (1 << j) ^ (1 << g)));
        }
      }
    }
  }
  return res;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  
  int n;
  cin >> n;
  Vec w(n), cnt(16);
  
  for (int i = 1, x, y, z; i < n; ++i) {
    cin >> x >> y >> z;
    w[x] ^= z;
    w[y] ^= z;
  }
  for (int i = 0; i < n; ++i) {
    ++cnt[w[i]];
  }

  int ans = 0, msk = 0;
  for (int i = 1; i < 16; ++i) {
    ans += cnt[i] / 2;
    cnt[i] %= 2;
    msk |= cnt[i] << i;
  }

  gay.resize(1 << 16);
  cout << ans + Dfs(msk) << endl;
  
  return 0;
}

### Prompt

Please formulate a cpp solution to 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 <bits/stdc++.h>

using namespace std;

typedef vector<int> Vec;

vector<int> gay;

int Dfs(int msk) {
  if (!msk) return 0;
  assert((msk & -msk) != msk);
  int &res = gay[msk];
  if (res) return res;
  res = 2233;
  for (int i = 0; i < 16; ++i) {
    if (msk >> i & 1) {
      for (int j = i + 1; j < 16; ++j) {
        if (msk >> j & 1) {
          int g = i ^ j;
          res = min(res, 1 + (msk >> g & 1) + Dfs(msk ^ (1 << i) ^ (1 << j) ^ (1 << g)));
        }
      }
    }
  }
  return res;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  
  int n;
  cin >> n;
  Vec w(n), cnt(16);
  
  for (int i = 1, x, y, z; i < n; ++i) {
    cin >> x >> y >> z;
    w[x] ^= z;
    w[y] ^= z;
  }
  for (int i = 0; i < n; ++i) {
    ++cnt[w[i]];
  }

  int ans = 0, msk = 0;
  for (int i = 1; i < 16; ++i) {
    ans += cnt[i] / 2;
    cnt[i] %= 2;
    msk |= cnt[i] << i;
  }

  gay.resize(1 << 16);
  cout << ans + Dfs(msk) << endl;
  
  return 0;
}

```