#include <bits/stdc++.h>

using namespace std;
const int N = 1e5 + 5, M = 16, inf = 1e9;
int a[N], cnt[M], p[M], dp[1<<M], Xor[1<<M];

int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n; cin >> n;
    for (int i = 0; i < n - 1; i++) {
        int u, v, x; cin >> u >> v >> x;
        a[u] ^= x, a[v] ^= x;
    }
    for (int i = 0; i < n; i++) cnt[a[i]]++;
    int res = cnt[0], m = 0;
    for (int i = 1; i < M; i++) {
        res += (cnt[i]>>1);
        if (cnt[i]&1) p[m++] = i;
    }
    fill(dp + 1, dp + (1<<M), -inf);
    for (int mask = 1; mask < (1<<m); mask++) {
        Xor[mask] = p[__builtin_ctz(mask)] ^ Xor[mask^(mask&(-mask))];
        for (int sub = mask; sub; sub = (sub - 1)&mask) if (Xor[sub] == 0)
            dp[mask] = max(dp[mask], dp[mask^sub] + 1);
    }
    cout << n - res - dp[(1<<m) - 1] << "\n";

    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 = 16, inf = 1e9;
int a[N], cnt[M], p[M], dp[1<<M], Xor[1<<M];

int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n; cin >> n;
    for (int i = 0; i < n - 1; i++) {
        int u, v, x; cin >> u >> v >> x;
        a[u] ^= x, a[v] ^= x;
    }
    for (int i = 0; i < n; i++) cnt[a[i]]++;
    int res = cnt[0], m = 0;
    for (int i = 1; i < M; i++) {
        res += (cnt[i]>>1);
        if (cnt[i]&1) p[m++] = i;
    }
    fill(dp + 1, dp + (1<<M), -inf);
    for (int mask = 1; mask < (1<<m); mask++) {
        Xor[mask] = p[__builtin_ctz(mask)] ^ Xor[mask^(mask&(-mask))];
        for (int sub = mask; sub; sub = (sub - 1)&mask) if (Xor[sub] == 0)
            dp[mask] = max(dp[mask], dp[mask^sub] + 1);
    }
    cout << n - res - dp[(1<<m) - 1] << "\n";

    return 0;
}

```