#include <bits/stdc++.h>
using namespace std;

int main() {
    int n; scanf("%d", &n);
    vector<int> a(n);
    for (int i = 0; i < n - 1; ++i) {
        int u, v, w; scanf("%d%d%d", &u, &v, &w);
        a[u] ^= w;
        a[v] ^= w;
    }
    vector<int> cnt(16);
    for (int i = 0; i < n; ++i) cnt[a[i]]++;
    int ans = 0;
    cnt[0] = 0;
    for (int i = 1; i < 16; ++i) {
        ans += cnt[i] / 2;
        cnt[i] &= 1;
    }
    vector<int> gd(1 << 16), dp(1 << 16);
    for (int s = 0; s < (1 << 16); ++s) {
        int v = 0;
        for (int i = 0; i < 16; ++i) {
            if (s >> i & 1) v ^= i;
        }
        if (v == 0) gd[s] = 1;
    }
    dp[0] = 0;
    for (int s = 1; s < (1 << 16); ++s) {
        dp[s] = -(1 << 16);
        for (int sub = s; sub > 0; sub = (sub - 1) & s) {
            if (gd[sub]) dp[s] = max(dp[s], dp[s ^ sub] + 1);
        }
    }
    int mask = 0;
    for (int i = 0; i < 16; ++i) {
        if (cnt[i] > 0) {
            mask |= (1 << i);
            ans++;
        }
    }
    printf("%d\n", ans - dp[mask]);
    return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;

int main() {
    int n; scanf("%d", &n);
    vector<int> a(n);
    for (int i = 0; i < n - 1; ++i) {
        int u, v, w; scanf("%d%d%d", &u, &v, &w);
        a[u] ^= w;
        a[v] ^= w;
    }
    vector<int> cnt(16);
    for (int i = 0; i < n; ++i) cnt[a[i]]++;
    int ans = 0;
    cnt[0] = 0;
    for (int i = 1; i < 16; ++i) {
        ans += cnt[i] / 2;
        cnt[i] &= 1;
    }
    vector<int> gd(1 << 16), dp(1 << 16);
    for (int s = 0; s < (1 << 16); ++s) {
        int v = 0;
        for (int i = 0; i < 16; ++i) {
            if (s >> i & 1) v ^= i;
        }
        if (v == 0) gd[s] = 1;
    }
    dp[0] = 0;
    for (int s = 1; s < (1 << 16); ++s) {
        dp[s] = -(1 << 16);
        for (int sub = s; sub > 0; sub = (sub - 1) & s) {
            if (gd[sub]) dp[s] = max(dp[s], dp[s ^ sub] + 1);
        }
    }
    int mask = 0;
    for (int i = 0; i < 16; ++i) {
        if (cnt[i] > 0) {
            mask |= (1 << i);
            ans++;
        }
    }
    printf("%d\n", ans - dp[mask]);
    return 0;
}

```