#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int MX=100010, inf=2e9;

int n, A[MX];
int cnt[16];

int B[1<<16];
int D[1<<16];

int d(int mask){
    int &res=D[mask];
    if(res<inf) return res;
    for(int i=mask; i>0; i=(i-1)&mask)
        if(B[i]>=0) res=min(res, d(mask^i)+B[i]-1);
    return res;
}

bool valid(int s){
    int res=0;
    for(int i=0; i<16; i++) if(s&(1<<i)) res^=i;
    return res==0;
}

int main(){
    ios::sync_with_stdio(0); cin.tie(0);
    cin>>n;
    for(int i=1; i<n; i++){
        int u,v,c; cin>>u>>v>>c;
        A[u]^=c, A[v]^=c;
    }

    for(int i=0; i<n; i++) cnt[A[i]]++;
    int ans=0, mask=0;
    for(int i=1; i<16; i++) ans+=cnt[i]/2, cnt[i]%=2;
    for(int i=1; i<16; i++) if(cnt[i]) mask|=(1<<i);

    for(int i=0; i<(1<<16); i++){
        if(valid(i)) B[i]=__builtin_popcount(i);
        else B[i]=-1;
    }

    for(int i=0; i<(1<<16); i++) D[i]=inf;
    D[0]=0;

    ans+=d(mask);

    cout<<ans<<'\n';

    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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int MX=100010, inf=2e9;

int n, A[MX];
int cnt[16];

int B[1<<16];
int D[1<<16];

int d(int mask){
    int &res=D[mask];
    if(res<inf) return res;
    for(int i=mask; i>0; i=(i-1)&mask)
        if(B[i]>=0) res=min(res, d(mask^i)+B[i]-1);
    return res;
}

bool valid(int s){
    int res=0;
    for(int i=0; i<16; i++) if(s&(1<<i)) res^=i;
    return res==0;
}

int main(){
    ios::sync_with_stdio(0); cin.tie(0);
    cin>>n;
    for(int i=1; i<n; i++){
        int u,v,c; cin>>u>>v>>c;
        A[u]^=c, A[v]^=c;
    }

    for(int i=0; i<n; i++) cnt[A[i]]++;
    int ans=0, mask=0;
    for(int i=1; i<16; i++) ans+=cnt[i]/2, cnt[i]%=2;
    for(int i=1; i<16; i++) if(cnt[i]) mask|=(1<<i);

    for(int i=0; i<(1<<16); i++){
        if(valid(i)) B[i]=__builtin_popcount(i);
        else B[i]=-1;
    }

    for(int i=0; i<(1<<16); i++) D[i]=inf;
    D[0]=0;

    ans+=d(mask);

    cout<<ans<<'\n';

    return 0;
}
```