#include<bits/stdc++.h>

using namespace std;

using int64 = long long;


int N, X[100000];
int dp[1 << 16];
int ok[1 << 16];

int rec(int bit) {
  if(~dp[bit]) return dp[bit];
  int ret = ok[bit];
  for(int i = bit; i > 0; i = (i - 1) & bit) {
    if(i == bit) continue;
    int other = i ^bit;
    ret = min(ret, rec(i) + rec(other));
  }
  return dp[bit] = ret;
}


int main() {
  cin >> N;
  for(int i = 1; i < N; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    X[x] ^= z;
    X[y] ^= z;
  }
  int bit = 0, ret = 0;
  for(int i = 0; i < N; i++) {
    if(X[i] == 0) continue;
    if(bit & (1 << X[i])) ++ret;
    bit ^= 1 << X[i];
  }
  memset(dp, -1, sizeof(dp));
  for(int i = 1; i < (1 << 16); i++) {
    ok[i] = ok[i & (i - 1)] + 1;
  }
  for(int i = 1; i < (1 << 16); i++) {
    int v = 0;
    for(int j = 0; j < 16; j++) {
      if((i >> j) & 1) v ^= j;
    }
    if(v == 0) --ok[i];
  }
  cout << ret + rec(bit) << endl;
}

### Prompt

Create a solution in cpp for 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;

using int64 = long long;


int N, X[100000];
int dp[1 << 16];
int ok[1 << 16];

int rec(int bit) {
  if(~dp[bit]) return dp[bit];
  int ret = ok[bit];
  for(int i = bit; i > 0; i = (i - 1) & bit) {
    if(i == bit) continue;
    int other = i ^bit;
    ret = min(ret, rec(i) + rec(other));
  }
  return dp[bit] = ret;
}


int main() {
  cin >> N;
  for(int i = 1; i < N; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    X[x] ^= z;
    X[y] ^= z;
  }
  int bit = 0, ret = 0;
  for(int i = 0; i < N; i++) {
    if(X[i] == 0) continue;
    if(bit & (1 << X[i])) ++ret;
    bit ^= 1 << X[i];
  }
  memset(dp, -1, sizeof(dp));
  for(int i = 1; i < (1 << 16); i++) {
    ok[i] = ok[i & (i - 1)] + 1;
  }
  for(int i = 1; i < (1 << 16); i++) {
    int v = 0;
    for(int j = 0; j < 16; j++) {
      if((i >> j) & 1) v ^= j;
    }
    if(v == 0) --ok[i];
  }
  cout << ret + rec(bit) << endl;
}

```