	//     . .. ... .... ..... be name khoda ..... .... ... .. .     \\

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

inline int in() { int x; scanf("%d", &x); return x; }
const int N = 200002, M = 16;

int a[N], cnt[M];
int dp[1 << M], val[1 << M];

int main()
{
	int n = in();
	for(int i = 1; i < n; i++)
	{
		int u = in(), v = in(), x = in();
		a[u] ^= x;
		a[v] ^= x;
	}
	for(int i = 0; i < n; i++)
		cnt[a[i]]++;
	vector <int> x;
	int reserved = cnt[0];
	cnt[0] = 0;
	for(int i = 0; i < n; i++)
	{
		if(cnt[a[i]] % 2)
			x.push_back(a[i]);
		reserved += cnt[a[i]]/2;
		cnt[a[i]] = 0;
	}
	int m = x.size();
	int tot = (1 << m);
	for(int i = 0; i < tot; i++)
		for(int j = 0; j < m; j++)
			if(i >> j & 1)
				val[i] ^= x[j];
	for(int mask = 1; mask < tot; mask++)
	{
		if(val[mask] != 0)
			continue;
		dp[mask] = 1;
		for(int smask = mask; smask > 0; smask = (smask - 1) & mask)
			if(!val[smask])
				dp[mask] = max(dp[mask], dp[smask] + dp[mask ^ smask]);
	}
	cout << n - dp[tot - 1] - reserved << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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
	//     . .. ... .... ..... be name khoda ..... .... ... .. .     \\

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

inline int in() { int x; scanf("%d", &x); return x; }
const int N = 200002, M = 16;

int a[N], cnt[M];
int dp[1 << M], val[1 << M];

int main()
{
	int n = in();
	for(int i = 1; i < n; i++)
	{
		int u = in(), v = in(), x = in();
		a[u] ^= x;
		a[v] ^= x;
	}
	for(int i = 0; i < n; i++)
		cnt[a[i]]++;
	vector <int> x;
	int reserved = cnt[0];
	cnt[0] = 0;
	for(int i = 0; i < n; i++)
	{
		if(cnt[a[i]] % 2)
			x.push_back(a[i]);
		reserved += cnt[a[i]]/2;
		cnt[a[i]] = 0;
	}
	int m = x.size();
	int tot = (1 << m);
	for(int i = 0; i < tot; i++)
		for(int j = 0; j < m; j++)
			if(i >> j & 1)
				val[i] ^= x[j];
	for(int mask = 1; mask < tot; mask++)
	{
		if(val[mask] != 0)
			continue;
		dp[mask] = 1;
		for(int smask = mask; smask > 0; smask = (smask - 1) & mask)
			if(!val[smask])
				dp[mask] = max(dp[mask], dp[smask] + dp[mask ^ smask]);
	}
	cout << n - dp[tot - 1] - reserved << endl;
}

```