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

int n;
int a[100005];
int cnt[16];
int ok[1 << 16], f[1 << 16];
int ans = 0;

int main() {
	ios_base::sync_with_stdio(false);
	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) {
		ans += cnt[a[i]] && a[i];
		cnt[a[i]] ^= 1;
	}
	int all = 0;
	for (int i = 1; i < 16; ++i) if (cnt[i]) {
		all |= (1 << i);
	}
	for (int mask = 0; mask < (1 << 16); ++mask) {
		int sum = 0;
		for (int i = 1; i < 16; ++i) if ((mask >> i) & 1) {
			sum ^= i;
		} 
		ok[mask] = sum == 0;
	} 
	for (int mask = 0; mask < (1 << 16); ++mask) {
		int cur = mask;
		while (cur) {
			if (ok[cur]) f[mask] = max(f[mask], f[mask ^ cur] + 1);
			cur = (cur - 1) & mask;
		}
	}
	cout << (ans + __builtin_popcount(all) - f[all]) << '\n';
}
### 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;

int n;
int a[100005];
int cnt[16];
int ok[1 << 16], f[1 << 16];
int ans = 0;

int main() {
	ios_base::sync_with_stdio(false);
	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) {
		ans += cnt[a[i]] && a[i];
		cnt[a[i]] ^= 1;
	}
	int all = 0;
	for (int i = 1; i < 16; ++i) if (cnt[i]) {
		all |= (1 << i);
	}
	for (int mask = 0; mask < (1 << 16); ++mask) {
		int sum = 0;
		for (int i = 1; i < 16; ++i) if ((mask >> i) & 1) {
			sum ^= i;
		} 
		ok[mask] = sum == 0;
	} 
	for (int mask = 0; mask < (1 << 16); ++mask) {
		int cur = mask;
		while (cur) {
			if (ok[cur]) f[mask] = max(f[mask], f[mask ^ cur] + 1);
			cur = (cur - 1) & mask;
		}
	}
	cout << (ans + __builtin_popcount(all) - f[all]) << '\n';
}
```