// KALAM
# include<bits/stdc++.h>

using namespace std;

const int N = 100000 + 77 , L = 16;
int n , A , a[N] , T[L];
int dp[1 << L] , XOR[1 << L];
vector < int > V;
int main() {
   scanf("%d" , & n);
   for(int v , u , x , i = 1;i < n;++ i)
      scanf("%d %d %d" , & v , & u , & x) , a[++ v] ^= x , a[++ u] ^= x;
   for(int i = 1;i <= n;++ i)
      ++ T[a[i]];
   T[0] = 1;
   for(int i = 1;i < L;++ i)
      A += (T[i] >> 1) , T[i] &= 1;
   for(int i = 1;i < L;++ i)
      if(T[i] > 0)
         V.push_back(i);

   int sz = V.size();
   for(int msk = 1;msk < (1 << sz);++ msk) {
      dp[msk] = N - 1;
      for(int i = 0;i < sz;++ i)
         if((1 << i) & msk)
            XOR[msk] ^= V[i];
      if(XOR[msk] > 0)
         continue ;
      dp[msk] = __builtin_popcount(msk) - 1;
      for(int sub = msk;sub > 0;sub = (sub - 1) & msk)
         if(XOR[sub] == 0)
            dp[msk] = min(dp[msk] , dp[sub] + dp[msk ^ sub]);
   }
   printf("%d\n" , A + dp[(1 << sz) - 1]);
   return 0;
}

### Prompt

Please create a solution in Cpp 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
// KALAM
# include<bits/stdc++.h>

using namespace std;

const int N = 100000 + 77 , L = 16;
int n , A , a[N] , T[L];
int dp[1 << L] , XOR[1 << L];
vector < int > V;
int main() {
   scanf("%d" , & n);
   for(int v , u , x , i = 1;i < n;++ i)
      scanf("%d %d %d" , & v , & u , & x) , a[++ v] ^= x , a[++ u] ^= x;
   for(int i = 1;i <= n;++ i)
      ++ T[a[i]];
   T[0] = 1;
   for(int i = 1;i < L;++ i)
      A += (T[i] >> 1) , T[i] &= 1;
   for(int i = 1;i < L;++ i)
      if(T[i] > 0)
         V.push_back(i);

   int sz = V.size();
   for(int msk = 1;msk < (1 << sz);++ msk) {
      dp[msk] = N - 1;
      for(int i = 0;i < sz;++ i)
         if((1 << i) & msk)
            XOR[msk] ^= V[i];
      if(XOR[msk] > 0)
         continue ;
      dp[msk] = __builtin_popcount(msk) - 1;
      for(int sub = msk;sub > 0;sub = (sub - 1) & msk)
         if(XOR[sub] == 0)
            dp[msk] = min(dp[msk] , dp[sub] + dp[msk ^ sub]);
   }
   printf("%d\n" , A + dp[(1 << sz) - 1]);
   return 0;
}

```