#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

const int N=100010;

int n,a[N],cnt[N],f[N],xs[N],lst[N];

int main(){
  scanf("%d",&n);
  for(int i=1,x,y,w;i<n;i++)
    scanf("%d%d%d",&x,&y,&w),a[x]^=w,a[y]^=w;
  for(int i=0;i<n;i++) cnt[a[i]]++;
  int ans=cnt[0]; cnt[0]=0;
  for(int i=1;i<16;i++) ans+=cnt[i]/2,cnt[i]%=2;
  for(int i=1;i<(1<<15);i++){
    lst[i]=(i&1)?1:lst[i>>1]+1;
    xs[i]=xs[i-(1<<lst[i]-1)]^lst[i];
    for(int j=i;j;j=(j-1)&i)
      if(xs[j]==0) f[i]=max(f[i],f[i^j]+1);
  }
  int S=0;
  for(int i=1;i<16;i++) S|=cnt[i]*(1<<i-1);
  ans+=f[S];
  printf("%d\n",n-ans);
  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 <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

const int N=100010;

int n,a[N],cnt[N],f[N],xs[N],lst[N];

int main(){
  scanf("%d",&n);
  for(int i=1,x,y,w;i<n;i++)
    scanf("%d%d%d",&x,&y,&w),a[x]^=w,a[y]^=w;
  for(int i=0;i<n;i++) cnt[a[i]]++;
  int ans=cnt[0]; cnt[0]=0;
  for(int i=1;i<16;i++) ans+=cnt[i]/2,cnt[i]%=2;
  for(int i=1;i<(1<<15);i++){
    lst[i]=(i&1)?1:lst[i>>1]+1;
    xs[i]=xs[i-(1<<lst[i]-1)]^lst[i];
    for(int j=i;j;j=(j-1)&i)
      if(xs[j]==0) f[i]=max(f[i],f[i^j]+1);
  }
  int S=0;
  for(int i=1;i<16;i++) S|=cnt[i]*(1<<i-1);
  ans+=f[S];
  printf("%d\n",n-ans);
  return 0;
}

```