#include <bits/stdc++.h>
using namespace std;
const int maxn=100005;
int n,a[maxn],cnt[16],dp[1<<16];
vector<int> v;
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n-1;++i) {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        a[u]^=w;
        a[v]^=w;
    }
    for (int i=0;i<n;++i)
        ++cnt[a[i]];
    int res=0;
    for (int i=1;i<16;++i) {
        res+=cnt[i]/2;
        if (cnt[i]&1)
            v.push_back(i);
    }
    dp[0]=0;
    for (int i=1;i<1<<(int)v.size();++i) {
        int mask=0;
        for (int j=0;j<(int)v.size();++j)
            if (1&(i>>j))
                mask^=v[j];
        for (int j=0;j<(int)v.size();++j)
            if (1&(i>>j))
                dp[i]=max(dp[i],dp[i^(1<<j)]+!mask);
    }
    return cout<<res+v.size()-dp[(1<<(int)v.size())-1]<<endl,0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int maxn=100005;
int n,a[maxn],cnt[16],dp[1<<16];
vector<int> v;
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n-1;++i) {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        a[u]^=w;
        a[v]^=w;
    }
    for (int i=0;i<n;++i)
        ++cnt[a[i]];
    int res=0;
    for (int i=1;i<16;++i) {
        res+=cnt[i]/2;
        if (cnt[i]&1)
            v.push_back(i);
    }
    dp[0]=0;
    for (int i=1;i<1<<(int)v.size();++i) {
        int mask=0;
        for (int j=0;j<(int)v.size();++j)
            if (1&(i>>j))
                mask^=v[j];
        for (int j=0;j<(int)v.size();++j)
            if (1&(i>>j))
                dp[i]=max(dp[i],dp[i^(1<<j)]+!mask);
    }
    return cout<<res+v.size()-dp[(1<<(int)v.size())-1]<<endl,0;
}

```