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

int n, sum[100005], cnt[16], dp[1 << 15];

int main() {
	scanf("%d", &n);
	for (int i = 1, u, v, c; i < n; ++i) {
		scanf("%d%d%d", &u, &v, &c);
		sum[u] ^= c, sum[v] ^= c;
	}
	for (int i = 0; i < n; ++i) ++cnt[sum[i]];
	int ans = 0, buc = 0;
	for (int i = 1; i < 16; ++i) ans += cnt[i] >> 1, buc |= (cnt[i] & 1) << i - 1;
	for (int s = 1; s <= buc; ++s) {
		int t = 0;
		for (int i = 0; i < 15; ++i)
			if (s >> i & 1) ++dp[s], t ^= i + 1;
		dp[s] -= !t;
		for (int t = s - 1 & s; t; t = t - 1 & s)
			dp[s] = min(dp[s], dp[t] + dp[s ^ t]);
	}
	printf("%d\n", ans + dp[buc]);
}

### Prompt

Your task is to create 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, sum[100005], cnt[16], dp[1 << 15];

int main() {
	scanf("%d", &n);
	for (int i = 1, u, v, c; i < n; ++i) {
		scanf("%d%d%d", &u, &v, &c);
		sum[u] ^= c, sum[v] ^= c;
	}
	for (int i = 0; i < n; ++i) ++cnt[sum[i]];
	int ans = 0, buc = 0;
	for (int i = 1; i < 16; ++i) ans += cnt[i] >> 1, buc |= (cnt[i] & 1) << i - 1;
	for (int s = 1; s <= buc; ++s) {
		int t = 0;
		for (int i = 0; i < 15; ++i)
			if (s >> i & 1) ++dp[s], t ^= i + 1;
		dp[s] -= !t;
		for (int t = s - 1 & s; t; t = t - 1 & s)
			dp[s] = min(dp[s], dp[t] + dp[s ^ t]);
	}
	printf("%d\n", ans + dp[buc]);
}

```