#include <bits/stdc++.h>

using namespace std;

const int MX = 100000;

int dp[1 << 16];

int g(int mask) {
	if (mask == 0) return 0;
	if (dp[mask] != 0) return dp[mask];
	
	dp[mask] = 42;
	int j = 0;
	while ((mask >> j) % 2 == 0) j++;
	for (int k = j + 1; (1 << k) <= mask; k++)
		if ((mask >> k) % 2 == 1) {
			int val = g(mask ^ (1 << j) ^ (1 << k) ^ (1 << (k ^ j))) + 1;
			if ((mask >> (k ^ j)) % 2 == 1) val++;
			dp[mask] = min(dp[mask], val);
		}
	
	return dp[mask];
}

int f[MX], cnt[16];

int main() {
	int n;
	ignore = scanf("%d", &n);
	for (int i = 1, u, v, a; i < n; i++) {
		ignore = scanf("%d %d %d", &u, &v, &a);
		f[u] ^= a;
		f[v] ^= a;
	}
	
	for (int i = 0; i < n; i++) cnt[f[i]]++;
	
	int ans = 0, mask = 0;
	for (int i = 1; i < 16; i++) {
		ans += cnt[i] / 2;
		mask |= (cnt[i] & 1) << i;
	}
	
	printf("%d\n", ans + g(mask));
	
	return 0;
}

### Prompt

Develop a solution in cpp 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;

const int MX = 100000;

int dp[1 << 16];

int g(int mask) {
	if (mask == 0) return 0;
	if (dp[mask] != 0) return dp[mask];
	
	dp[mask] = 42;
	int j = 0;
	while ((mask >> j) % 2 == 0) j++;
	for (int k = j + 1; (1 << k) <= mask; k++)
		if ((mask >> k) % 2 == 1) {
			int val = g(mask ^ (1 << j) ^ (1 << k) ^ (1 << (k ^ j))) + 1;
			if ((mask >> (k ^ j)) % 2 == 1) val++;
			dp[mask] = min(dp[mask], val);
		}
	
	return dp[mask];
}

int f[MX], cnt[16];

int main() {
	int n;
	ignore = scanf("%d", &n);
	for (int i = 1, u, v, a; i < n; i++) {
		ignore = scanf("%d %d %d", &u, &v, &a);
		f[u] ^= a;
		f[v] ^= a;
	}
	
	for (int i = 0; i < n; i++) cnt[f[i]]++;
	
	int ans = 0, mask = 0;
	for (int i = 1; i < 16; i++) {
		ans += cnt[i] / 2;
		mask |= (cnt[i] & 1) << i;
	}
	
	printf("%d\n", ans + g(mask));
	
	return 0;
}

```