#include<bits/stdc++.h>
using namespace std;

int n,ans;
int f[100005];
int v[16],cnt,dp[1<<15],x[1<<15];

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n;
    for(int i=1;i<n;i++){
        int a,b,c;
        cin>>a>>b>>c;
        f[a]^=c;
        f[b]^=c;
    }
    for(int i=0;i<n;i++){
        if(!f[i])ans++;
        else v[f[i]]++;
    }
    for(int i=1;i<=15;i++){
        ans+=v[i]>>1;
        v[i]&=1;
        if(v[i])v[cnt++]=i;
    }
    for(int s=0;s<1<<cnt;s++){
        for(int i=0;i<cnt;i++){
            if(s>>i&1)x[s]^=v[i];
        }
    }
    for(int s=0;s<1<<cnt;s++){
        int rs=s^((1<<cnt)-1);
        for(int t=rs;t;t=(t-1)&rs){
            if(!x[t])dp[s|t]=max(dp[s|t],dp[s]+1);
        }
    }
    ans+=dp[(1<<cnt)-1];
    cout<<n-ans<<endl;

    return 0;
}
### Prompt

Generate 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<bits/stdc++.h>
using namespace std;

int n,ans;
int f[100005];
int v[16],cnt,dp[1<<15],x[1<<15];

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n;
    for(int i=1;i<n;i++){
        int a,b,c;
        cin>>a>>b>>c;
        f[a]^=c;
        f[b]^=c;
    }
    for(int i=0;i<n;i++){
        if(!f[i])ans++;
        else v[f[i]]++;
    }
    for(int i=1;i<=15;i++){
        ans+=v[i]>>1;
        v[i]&=1;
        if(v[i])v[cnt++]=i;
    }
    for(int s=0;s<1<<cnt;s++){
        for(int i=0;i<cnt;i++){
            if(s>>i&1)x[s]^=v[i];
        }
    }
    for(int s=0;s<1<<cnt;s++){
        int rs=s^((1<<cnt)-1);
        for(int t=rs;t;t=(t-1)&rs){
            if(!x[t])dp[s|t]=max(dp[s|t],dp[s]+1);
        }
    }
    ans+=dp[(1<<cnt)-1];
    cout<<n-ans<<endl;

    return 0;
}
```