#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

#define N 100010
#define M 65540

int n,s[N],x,ans,u,v,w;
int f[M],inf,sum[20]; 

int dfs(int S) {
    if(!S)return 0; 
    if(f[S]<inf)return f[S];
    for(int i=0;i<16;i++)if((S>>i)&1)
        for(int j=0;j<16;j++)if(i!=j&&(S>>j)&1) {
        int p=i^j,x=S^(1<<i)^(1<<j)^(1<<p);
        if(S>>p&1)f[S]=min(f[S],dfs(x)+2);
        else f[S]=min(f[S],dfs(x)+1);
    }
    return f[S];
}

int main() {
    scanf("%d",&n); 
    for(int i = 1 ; i < n ; i++) {
        scanf("%d%d%d", &u,&v,&w);
        s[u] ^= w, s[v] ^= w; 
    }
    for(int i = 0 ; i < n ; i++) sum[s[i]]++; 
    for(int i = 1 ; i < 16 ; i++) {
        ans += (sum[i] >> 1);
        x += (1 << i) * (sum[i] & 1); 
    }
    memset(f, 127, sizeof(f)); 
    inf = f[0]; 
    printf("%d\n", ans + dfs(x)); 
    //system("pause");
    return 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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

#define N 100010
#define M 65540

int n,s[N],x,ans,u,v,w;
int f[M],inf,sum[20]; 

int dfs(int S) {
    if(!S)return 0; 
    if(f[S]<inf)return f[S];
    for(int i=0;i<16;i++)if((S>>i)&1)
        for(int j=0;j<16;j++)if(i!=j&&(S>>j)&1) {
        int p=i^j,x=S^(1<<i)^(1<<j)^(1<<p);
        if(S>>p&1)f[S]=min(f[S],dfs(x)+2);
        else f[S]=min(f[S],dfs(x)+1);
    }
    return f[S];
}

int main() {
    scanf("%d",&n); 
    for(int i = 1 ; i < n ; i++) {
        scanf("%d%d%d", &u,&v,&w);
        s[u] ^= w, s[v] ^= w; 
    }
    for(int i = 0 ; i < n ; i++) sum[s[i]]++; 
    for(int i = 1 ; i < 16 ; i++) {
        ans += (sum[i] >> 1);
        x += (1 << i) * (sum[i] & 1); 
    }
    memset(f, 127, sizeof(f)); 
    inf = f[0]; 
    printf("%d\n", ans + dfs(x)); 
    //system("pause");
    return 0; 
}
```