#include<bits/stdc++.h>
#define N 100005
#define M 65540
using namespace std;
int n,s[N],S,ans,f[M],inf,sum[20];

int dfs(int S)
{
  if(!S)return 0; 
  if(f[S]<inf)return f[S];
  for(int i=0;i<16;i++)if((S>>i)&1)
    for(int j=0;j<16;j++)if(i!=j&&(S>>j)&1)
    {
      int p=i^j,x=S^(1<<i)^(1<<j)^(1<<p);
      if(S>>p&1)f[S]=min(f[S],dfs(x)+2);
      else f[S]=min(f[S],dfs(x)+1);
      //枚举每次异或的两个节点i,j,i异或掉j,j异或掉j,还剩下p=i^j
      //如果p本身存在那两个p直接消掉 
    }
  return f[S];
}

int main()
{
  int a,b,c;
  scanf("%d",&n);
  for(int i=1;i<n;i++)
    scanf("%d%d%d",&a,&b,&c),s[a]^=c,s[b]^=c;
  for(int i=0;i<n;i++)sum[s[i]]++;
  for(int i=1;i<16;i++)
    ans+=(sum[i]>>1),S+=(1<<i)*(sum[i]&1);
  memset(f,127,sizeof(f));inf=f[0];
  printf("%d\n",ans+dfs(S));
  return 0;
}
### Prompt

In Cpp, your task is to solve 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>
#define N 100005
#define M 65540
using namespace std;
int n,s[N],S,ans,f[M],inf,sum[20];

int dfs(int S)
{
  if(!S)return 0; 
  if(f[S]<inf)return f[S];
  for(int i=0;i<16;i++)if((S>>i)&1)
    for(int j=0;j<16;j++)if(i!=j&&(S>>j)&1)
    {
      int p=i^j,x=S^(1<<i)^(1<<j)^(1<<p);
      if(S>>p&1)f[S]=min(f[S],dfs(x)+2);
      else f[S]=min(f[S],dfs(x)+1);
      //枚举每次异或的两个节点i,j,i异或掉j,j异或掉j,还剩下p=i^j
      //如果p本身存在那两个p直接消掉 
    }
  return f[S];
}

int main()
{
  int a,b,c;
  scanf("%d",&n);
  for(int i=1;i<n;i++)
    scanf("%d%d%d",&a,&b,&c),s[a]^=c,s[b]^=c;
  for(int i=0;i<n;i++)sum[s[i]]++;
  for(int i=1;i<16;i++)
    ans+=(sum[i]>>1),S+=(1<<i)*(sum[i]&1);
  memset(f,127,sizeof(f));inf=f[0];
  printf("%d\n",ans+dfs(S));
  return 0;
}
```