#include <bits/stdc++.h> 
using namespace std; 
const int maxn = 1e5 + 100; 
const int maxm = 1 << 15 + 10; 
const int inf = 0x3f3f3f3f; 

int n, ans, state; 
int v[maxn], cnt[20], f[maxm]; 

int dp(int s) {
    if (!s) return 0; 
    if (f[s] < inf) return f[s]; 
    for (int i = 0; i <= 15; ++i) {
        if (!(s & (1 << i))) continue; 
        for (int j = 0; j <= 15; ++j) {
            if (i == j || !(s & (1 << j))) continue; 
            int x = i ^ j; 
            int y = s ^ (1 << i) ^ (1 << j) ^ (1 << x); 
            if (s & (1 << x)) f[s] = min(f[s], dp(y) + 2); 
            else f[s] = min(f[s], dp(y) + 1); 
        }
    }
    return f[s]; 
}

int main() {
    scanf("%d", &n); 
    for (int i = 1; i < n; ++i) {
        int x, y, z; 
        scanf("%d%d%d", &x, &y, &z); 
        x += 1, y += 1; 
        v[x] ^= z; v[y] ^= z; 
    }
    for (int i = 1; i <= n; ++i) cnt[v[i]]++; 
    for (int i = 1; i <= 15; ++i) {
        ans += (cnt[i] >> 1); 
        state += (cnt[i] & 1) * (1 << i); 
    }
    memset(f, 0x3f, sizeof(f)); 
    printf("%d\n", ans + dp(state)); 
    return 0; 
}
### Prompt

Your challenge is to write 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; 
const int maxn = 1e5 + 100; 
const int maxm = 1 << 15 + 10; 
const int inf = 0x3f3f3f3f; 

int n, ans, state; 
int v[maxn], cnt[20], f[maxm]; 

int dp(int s) {
    if (!s) return 0; 
    if (f[s] < inf) return f[s]; 
    for (int i = 0; i <= 15; ++i) {
        if (!(s & (1 << i))) continue; 
        for (int j = 0; j <= 15; ++j) {
            if (i == j || !(s & (1 << j))) continue; 
            int x = i ^ j; 
            int y = s ^ (1 << i) ^ (1 << j) ^ (1 << x); 
            if (s & (1 << x)) f[s] = min(f[s], dp(y) + 2); 
            else f[s] = min(f[s], dp(y) + 1); 
        }
    }
    return f[s]; 
}

int main() {
    scanf("%d", &n); 
    for (int i = 1; i < n; ++i) {
        int x, y, z; 
        scanf("%d%d%d", &x, &y, &z); 
        x += 1, y += 1; 
        v[x] ^= z; v[y] ^= z; 
    }
    for (int i = 1; i <= n; ++i) cnt[v[i]]++; 
    for (int i = 1; i <= 15; ++i) {
        ans += (cnt[i] >> 1); 
        state += (cnt[i] & 1) * (1 << i); 
    }
    memset(f, 0x3f, sizeof(f)); 
    printf("%d\n", ans + dp(state)); 
    return 0; 
}
```