#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

Create a solution in CPP for 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 <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; 
}
```