#include <bits/stdc++.h>

typedef long long ll;

using namespace std;

const int N = 595959;
const int K = 16;

int n, ans;
int v[N];
int cnt[N];
vector<int> vec;
int dp[1 << K];

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin >> n;
    for (int i = 0; i < n - 1; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        v[a] ^= c, v[b] ^= c;
    }
    for (int i = 0; i < n; i++)
    {
        cnt[v[i]]++;
    }
    ans += cnt[0];
    cnt[0] = 0;
    for (int i = 0; i < 16; i++)
    {
        ans += cnt[i] / 2;
        if (cnt[i] % 2) vec.push_back(i);
    }
    //fill(dp, dp + (1 << vec.size()), N);
    dp[0] = 0;
    for (int mask = 1; mask < (1 << vec.size()); mask++)
    {
        for (int submask = mask; submask; submask = mask & (submask - 1))
        {
            int x = 0;
            for (int i = 0; i < vec.size(); i++) if ((1 << i) & submask) x ^= vec[i];
            if (x == 0) dp[mask] = max(dp[mask], dp[mask - submask] + 1);
        }
    }
    cout << n - ans - dp[(1 << vec.size()) - 1];
}

### 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>

typedef long long ll;

using namespace std;

const int N = 595959;
const int K = 16;

int n, ans;
int v[N];
int cnt[N];
vector<int> vec;
int dp[1 << K];

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin >> n;
    for (int i = 0; i < n - 1; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        v[a] ^= c, v[b] ^= c;
    }
    for (int i = 0; i < n; i++)
    {
        cnt[v[i]]++;
    }
    ans += cnt[0];
    cnt[0] = 0;
    for (int i = 0; i < 16; i++)
    {
        ans += cnt[i] / 2;
        if (cnt[i] % 2) vec.push_back(i);
    }
    //fill(dp, dp + (1 << vec.size()), N);
    dp[0] = 0;
    for (int mask = 1; mask < (1 << vec.size()); mask++)
    {
        for (int submask = mask; submask; submask = mask & (submask - 1))
        {
            int x = 0;
            for (int i = 0; i < vec.size(); i++) if ((1 << i) & submask) x ^= vec[i];
            if (x == 0) dp[mask] = max(dp[mask], dp[mask - submask] + 1);
        }
    }
    cout << n - ans - dp[(1 << vec.size()) - 1];
}

```