#include <algorithm>
#include <cstring>
#include <cstdio>
int val[100005], app[25], seq[25];
int dp[100005], pop[100005], sum[100005];
int main()
{
	// freopen("E.in", "r", stdin);
	int n, m = 0, ans = 0;
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		val[u] ^= w;
		val[v] ^= w;
	}
	for (int i = 0; i < n; i++)
		app[val[i]]++;
	for (int i = 1; i < 16; i++)
	{
		ans += app[i] >> 1;
		if (app[i] & 1)
			seq[m++] = i;
	}
	for (int i = 1; i < 1 << m; i++)
	{
		pop[i] = pop[i ^ (i & -i)] + 1;
		for (int j = 0; j < m; j++)
		{
			if (i & (1 << j))
				sum[i] ^= seq[j];
		}
	}
	memset(dp, 0x3f, sizeof(dp));
	dp[0] = 0;
	for (int i = 0; i + 1 < 1 << m; i++)
	{
		int rem = (1 << m) - 1 ^ i, low = rem & -rem;
		for (int j = rem ^ low; ; j = j - 1 & (rem ^ low))
		{
			int add = j | low;
			if (!sum[add])
				dp[i | add] = std::min(dp[i | add], dp[i] + pop[j]);
			if (!j)
				break;
		}
	}
	printf("%d\n", ans + dp[(1 << m) - 1]);
	return 0;
}

### 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 <algorithm>
#include <cstring>
#include <cstdio>
int val[100005], app[25], seq[25];
int dp[100005], pop[100005], sum[100005];
int main()
{
	// freopen("E.in", "r", stdin);
	int n, m = 0, ans = 0;
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		val[u] ^= w;
		val[v] ^= w;
	}
	for (int i = 0; i < n; i++)
		app[val[i]]++;
	for (int i = 1; i < 16; i++)
	{
		ans += app[i] >> 1;
		if (app[i] & 1)
			seq[m++] = i;
	}
	for (int i = 1; i < 1 << m; i++)
	{
		pop[i] = pop[i ^ (i & -i)] + 1;
		for (int j = 0; j < m; j++)
		{
			if (i & (1 << j))
				sum[i] ^= seq[j];
		}
	}
	memset(dp, 0x3f, sizeof(dp));
	dp[0] = 0;
	for (int i = 0; i + 1 < 1 << m; i++)
	{
		int rem = (1 << m) - 1 ^ i, low = rem & -rem;
		for (int j = rem ^ low; ; j = j - 1 & (rem ^ low))
		{
			int add = j | low;
			if (!sum[add])
				dp[i | add] = std::min(dp[i | add], dp[i] + pop[j]);
			if (!j)
				break;
		}
	}
	printf("%d\n", ans + dp[(1 << m) - 1]);
	return 0;
}

```