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

const int N = 100005;
const int MAX = 1 << 16;
const int inf = 1e9;

int n, a[N];
int cnt[16];
int dp[MAX];
int fxor[MAX];
int ans;
vector<int> rem;

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n;
	for (int i = 1; i < n; ++i) {
		int u, v, w; cin >> u >> v >> w;
		a[u] ^= w;
		a[v] ^= w;
	}

	for (int i = 0; i < n; ++i) cnt[a[i]]++;

	for (int i = 1; i < 16; ++i) {
		ans += cnt[i] / 2;
		cnt[i] %= 2;
		if (cnt[i]) rem.push_back(i);
	}

	int sz = rem.size();
	for (int mask = 0; mask < (1 << sz); ++mask) {
		for (int i = 0; i < sz; ++i) if (mask >> i & 1) fxor[mask] ^= rem[i];
	}
	
	for (int mask = 0; mask < (1 << sz); ++mask) dp[mask] = inf;
	dp[0] = 0;
	for (int mask = 1; mask < (1 << sz); ++mask) {
		for (int submask = mask; submask > 0; submask = mask & (submask - 1)) {
			if (fxor[submask] == 0) {
				dp[mask] = min(dp[mask], dp[mask ^ submask] + __builtin_popcount(submask) - 1);
			}
		}
	}

	cout << ans + dp[(1 << sz) - 1] << endl;
}
### Prompt

In cpp, your task is to solve 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 = 100005;
const int MAX = 1 << 16;
const int inf = 1e9;

int n, a[N];
int cnt[16];
int dp[MAX];
int fxor[MAX];
int ans;
vector<int> rem;

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n;
	for (int i = 1; i < n; ++i) {
		int u, v, w; cin >> u >> v >> w;
		a[u] ^= w;
		a[v] ^= w;
	}

	for (int i = 0; i < n; ++i) cnt[a[i]]++;

	for (int i = 1; i < 16; ++i) {
		ans += cnt[i] / 2;
		cnt[i] %= 2;
		if (cnt[i]) rem.push_back(i);
	}

	int sz = rem.size();
	for (int mask = 0; mask < (1 << sz); ++mask) {
		for (int i = 0; i < sz; ++i) if (mask >> i & 1) fxor[mask] ^= rem[i];
	}
	
	for (int mask = 0; mask < (1 << sz); ++mask) dp[mask] = inf;
	dp[0] = 0;
	for (int mask = 1; mask < (1 << sz); ++mask) {
		for (int submask = mask; submask > 0; submask = mask & (submask - 1)) {
			if (fxor[submask] == 0) {
				dp[mask] = min(dp[mask], dp[mask ^ submask] + __builtin_popcount(submask) - 1);
			}
		}
	}

	cout << ans + dp[(1 << sz) - 1] << endl;
}
```