#include <bits/stdc++.h> 
using namespace std; 
const int N = 1e5 + 5, M = (1 << 15);
int val[N], cnt[N], sum[M], f[M];
int main() {
	int n; scanf("%d", &n); 
    for (int i = 1; i < n; ++i) {
        int x, y, z; scanf("%d%d%d", &x, &y, &z); 
        val[++x] ^= z, val[++y] ^= z; 
    }
    for (int i = 1; i <= n; ++i) cnt[val[i]]++; 
    int ans = n - cnt[0], mask = 0;
    for (int i = 1; i <= 15; ++i) {
        ans -= cnt[i] / 2 ; 
        mask |= (cnt[i] & 1) << (i - 1); 
    }
    for (int i = 0; i < (1 << 15); ++i) 
        for (int j = 1; j <= 15; ++j) 
            if (i & (1 << (j - 1))) sum[i] ^= j; 
    for (int i = 1; i < (1 << 15); ++i) 
        for (int j = i; j; j = (j - 1) & i) 
            if (!sum[i]) f[i] = max(f[i], f[i ^ j] + 1); 
    printf("%d\n", ans - f[mask]); 
    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 N = 1e5 + 5, M = (1 << 15);
int val[N], cnt[N], sum[M], f[M];
int main() {
	int n; scanf("%d", &n); 
    for (int i = 1; i < n; ++i) {
        int x, y, z; scanf("%d%d%d", &x, &y, &z); 
        val[++x] ^= z, val[++y] ^= z; 
    }
    for (int i = 1; i <= n; ++i) cnt[val[i]]++; 
    int ans = n - cnt[0], mask = 0;
    for (int i = 1; i <= 15; ++i) {
        ans -= cnt[i] / 2 ; 
        mask |= (cnt[i] & 1) << (i - 1); 
    }
    for (int i = 0; i < (1 << 15); ++i) 
        for (int j = 1; j <= 15; ++j) 
            if (i & (1 << (j - 1))) sum[i] ^= j; 
    for (int i = 1; i < (1 << 15); ++i) 
        for (int j = i; j; j = (j - 1) & i) 
            if (!sum[i]) f[i] = max(f[i], f[i ^ j] + 1); 
    printf("%d\n", ans - f[mask]); 
    return 0; 
}

```