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

#define pii pair<int,int>
#define pll pair<ll,ll>
#define eb emplace_back
#define ll long long
#define nl '\n'
#define deb(x) cerr<<#x" = "<<x<<nl
#define in() ( { int a ; scanf("%d",&a); a; } )

const int N = 3e5 + 9;
const int mod = 1e9 + 7;

int a[N], cnt[16], dp[1<<16], pos[1<<16];
int32_t main()
{
    int n=in();
    for(int i=1; i<n; i++){
        int u=in(), v=in(), k=in();
        u++; v++;
        a[u]^=k; a[v]^=k;
    }
    int ans=0;
    for(int i=1; i<=n; i++) cnt[a[i]]++;
    ans+=cnt[0];
    for(int i=1; i<16; i++) ans+=cnt[i]>>1, cnt[i]%=2;
    cnt[0]=0;
    dp[0]=0;
    vector<int> v;
    for(int i=0; i<16; i++) if(cnt[i]) v.eb(i);
    int m=v.size();
    for(int mask=1; mask<(1<<m); mask++){
        int nw=0;
        for(int i=0; i<m; i++) if((mask>>i)&1) nw^=v[i];
        if(nw==0) pos[mask]=1;
    }
    for(int mask=1; mask<(1<<m); mask++){
        for(int sub=mask; sub; sub=(sub-1)&mask){
            if(pos[sub]) dp[mask]=max(dp[mask], dp[mask^sub]+1);
        }
    }
    ans+=dp[(1<<m)-1];
    cout<<n-ans<<nl;
    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;

#define pii pair<int,int>
#define pll pair<ll,ll>
#define eb emplace_back
#define ll long long
#define nl '\n'
#define deb(x) cerr<<#x" = "<<x<<nl
#define in() ( { int a ; scanf("%d",&a); a; } )

const int N = 3e5 + 9;
const int mod = 1e9 + 7;

int a[N], cnt[16], dp[1<<16], pos[1<<16];
int32_t main()
{
    int n=in();
    for(int i=1; i<n; i++){
        int u=in(), v=in(), k=in();
        u++; v++;
        a[u]^=k; a[v]^=k;
    }
    int ans=0;
    for(int i=1; i<=n; i++) cnt[a[i]]++;
    ans+=cnt[0];
    for(int i=1; i<16; i++) ans+=cnt[i]>>1, cnt[i]%=2;
    cnt[0]=0;
    dp[0]=0;
    vector<int> v;
    for(int i=0; i<16; i++) if(cnt[i]) v.eb(i);
    int m=v.size();
    for(int mask=1; mask<(1<<m); mask++){
        int nw=0;
        for(int i=0; i<m; i++) if((mask>>i)&1) nw^=v[i];
        if(nw==0) pos[mask]=1;
    }
    for(int mask=1; mask<(1<<m); mask++){
        for(int sub=mask; sub; sub=(sub-1)&mask){
            if(pos[sub]) dp[mask]=max(dp[mask], dp[mask^sub]+1);
        }
    }
    ans+=dp[(1<<m)-1];
    cout<<n-ans<<nl;
    return 0;
}

```