#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
const int N=1e5+5;
int n,a[N],vis[16],dp[1<<15],sum[1<<15];
vector<int>v;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int u,v,w;scanf("%d%d%d",&u,&v,&w);
        a[u]^=w;a[v]^=w;
    }
    for(int i=1;i<=n;i++)
        vis[a[i]]++;
    int ans=0;
    for(int i=1;i<=15;i++)
    {
        ans+=vis[i]/2;
        if(vis[i]&1)
            v.push_back(i);
    }
    memset(dp,inf,sizeof(dp));
    dp[0]=0;
    int m=v.size();
    int up=1<<m;
    for(int i=0;i<up;i++)
    {
        for(int j=0;j<m;j++)
            if(!(i>>j&1))
        {
            int st=i^(1<<j);
            sum[st]=sum[i]^v[j];
            if(sum[st]==0) dp[st]=min(dp[st],dp[i]);
            else dp[st]=min(dp[st],dp[i]+1);
        }
    }
    printf("%d\n",ans+dp[up-1]);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 inf 0x3f3f3f3f
using namespace std;
const int N=1e5+5;
int n,a[N],vis[16],dp[1<<15],sum[1<<15];
vector<int>v;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int u,v,w;scanf("%d%d%d",&u,&v,&w);
        a[u]^=w;a[v]^=w;
    }
    for(int i=1;i<=n;i++)
        vis[a[i]]++;
    int ans=0;
    for(int i=1;i<=15;i++)
    {
        ans+=vis[i]/2;
        if(vis[i]&1)
            v.push_back(i);
    }
    memset(dp,inf,sizeof(dp));
    dp[0]=0;
    int m=v.size();
    int up=1<<m;
    for(int i=0;i<up;i++)
    {
        for(int j=0;j<m;j++)
            if(!(i>>j&1))
        {
            int st=i^(1<<j);
            sum[st]=sum[i]^v[j];
            if(sum[st]==0) dp[st]=min(dp[st],dp[i]);
            else dp[st]=min(dp[st],dp[i]+1);
        }
    }
    printf("%d\n",ans+dp[up-1]);
}

```