#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;const int N=2*1e5+10;
int w[N];int d[N];int cnt[20];int res;int n;int st;bool book[N];int sxr[N]; 
int main()
{
    scanf("%d",&n);
    for(int i=1,u,v,va;i<n;i++){scanf("%d%d%d",&u,&v,&va);w[u]^=va;w[v]^=va;}
    for(int i=0;i<n;i++){cnt[w[i]]++;}
    for(int i=1;i<=15;i++)res+=cnt[i]/2,st|=(cnt[i]&1)<<(i-1);	
    for(int i=1;i<(1<<15);i++)d[i]=d[i>>1]+(i&1);
    for(int i=1;i<(1<<15);i++)d[i]-=1;
    for(int i=1;i<(1<<15);i++)
        for(int j=0;j<15;j++)if((i>>j)&1)sxr[i]^=(j+1);
    for(int i=1;i<(1<<15);i++) 
    {
        if(sxr[i]!=0)continue;
        for(int k=(i-1)&i;k;k=(k-1)&i)
            if(sxr[k]==0)d[i]=min(d[i],d[k]+d[i^k]);
    }printf("%d",res+d[st]);return 0;
}
### Prompt

Your challenge is to write 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<algorithm>
#include<queue>
using namespace std;const int N=2*1e5+10;
int w[N];int d[N];int cnt[20];int res;int n;int st;bool book[N];int sxr[N]; 
int main()
{
    scanf("%d",&n);
    for(int i=1,u,v,va;i<n;i++){scanf("%d%d%d",&u,&v,&va);w[u]^=va;w[v]^=va;}
    for(int i=0;i<n;i++){cnt[w[i]]++;}
    for(int i=1;i<=15;i++)res+=cnt[i]/2,st|=(cnt[i]&1)<<(i-1);	
    for(int i=1;i<(1<<15);i++)d[i]=d[i>>1]+(i&1);
    for(int i=1;i<(1<<15);i++)d[i]-=1;
    for(int i=1;i<(1<<15);i++)
        for(int j=0;j<15;j++)if((i>>j)&1)sxr[i]^=(j+1);
    for(int i=1;i<(1<<15);i++) 
    {
        if(sxr[i]!=0)continue;
        for(int k=(i-1)&i;k;k=(k-1)&i)
            if(sxr[k]==0)d[i]=min(d[i],d[k]+d[i^k]);
    }printf("%d",res+d[st]);return 0;
}
```