# include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;

int n, ans = 0, state = 0;
int a[MAXN], sum[MAXN], cnt[MAXN], f[MAXN];

int main()
{
    scanf("%d", &n);
    int u, v, w;
    for(int i = 1; i < n; i++) {
        scanf("%d%d%d", &u, &v, &w);
        a[u] ^= w, a[v] ^= w;
    }
    for (int i = 0; i < n; i++) ++cnt[a[i]];
    for (int i = 1; i <= 15; i++) ans += (cnt[i] >> 1), state |= ((cnt[i] & 1) << (i - 1));
    int sta_max = (1 << 15) - 1;
    for (int sta = 1; sta <= sta_max; sta++) f[sta] = f[sta & (sta - 1)] + 1;
    for (int sta = 1; sta <= sta_max; sta++) --f[sta];
    for (int sta = 1; sta <= sta_max; sta++)
        for (int i = 1; i <= 15; i++)
            if((sta >> (i - 1)) & 1) sum[sta] ^= i;
    for (int sta = 1; sta <= sta_max; sta++) {
        if(sum[sta]) continue;
        for (int i = (sta & (sta - 1)); i; i = ((i - 1) & sta))
            if (!sum[i]) f[sta] = min(f[sta], f[i] + f[sta ^ i]);
    }
    printf("%d\n", ans + f[state]);
    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 <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;

int n, ans = 0, state = 0;
int a[MAXN], sum[MAXN], cnt[MAXN], f[MAXN];

int main()
{
    scanf("%d", &n);
    int u, v, w;
    for(int i = 1; i < n; i++) {
        scanf("%d%d%d", &u, &v, &w);
        a[u] ^= w, a[v] ^= w;
    }
    for (int i = 0; i < n; i++) ++cnt[a[i]];
    for (int i = 1; i <= 15; i++) ans += (cnt[i] >> 1), state |= ((cnt[i] & 1) << (i - 1));
    int sta_max = (1 << 15) - 1;
    for (int sta = 1; sta <= sta_max; sta++) f[sta] = f[sta & (sta - 1)] + 1;
    for (int sta = 1; sta <= sta_max; sta++) --f[sta];
    for (int sta = 1; sta <= sta_max; sta++)
        for (int i = 1; i <= 15; i++)
            if((sta >> (i - 1)) & 1) sum[sta] ^= i;
    for (int sta = 1; sta <= sta_max; sta++) {
        if(sum[sta]) continue;
        for (int i = (sta & (sta - 1)); i; i = ((i - 1) & sta))
            if (!sum[i]) f[sta] = min(f[sta], f[i] + f[sta ^ i]);
    }
    printf("%d\n", ans + f[state]);
    return 0;
}
```