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

#define ll long long
#define pb push_back
#define bit(a , b) (((a)>>(b))&1)

const int maxn = 1e5 + 20;
const int maxm = 16;

int t[maxn] , a[maxm] , dp[1 << maxm];

int And[1 << maxm];

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n;
	cin >> n;

	for(int i = 0; i < n - 1; i++)
	{
		int a , b , c;
		cin >> a >> b >> c;

		t[a] ^= c;
		t[b] ^= c;
	}

	for(int i = 0; i < n; i++)
		a[t[i]]++;

	int res = a[0] , mask = 0;
	for(int i = 1; i < maxm; i++)
	{
		res += a[i] / 2;
		a[i] %= 2;

		mask += (1 << i) * a[i];
	}

	for(int i = 0; i < maxm; i++)
		And[1 << i] = i;

	for(int mask = 1; mask < (1 << maxm); mask++)
	{
		int lb = (mask&(-mask));

		if(lb == mask)
			continue;

		And[mask] = And[mask ^ lb] ^ And[lb];
	}

	for(int mask = 1; mask < (1 << maxm); mask++)
	{
		if(And[mask] != 0)
			continue;

		dp[mask] = 1;
		for(int submask = mask; submask > 0; submask = (submask - 1) & mask)
			if(submask != mask && And[submask] == 0)
				dp[mask] = max(dp[mask] , dp[submask] + 1);
	}

	cout << n - res - dp[mask] << endl;
}









### 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;

#define ll long long
#define pb push_back
#define bit(a , b) (((a)>>(b))&1)

const int maxn = 1e5 + 20;
const int maxm = 16;

int t[maxn] , a[maxm] , dp[1 << maxm];

int And[1 << maxm];

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n;
	cin >> n;

	for(int i = 0; i < n - 1; i++)
	{
		int a , b , c;
		cin >> a >> b >> c;

		t[a] ^= c;
		t[b] ^= c;
	}

	for(int i = 0; i < n; i++)
		a[t[i]]++;

	int res = a[0] , mask = 0;
	for(int i = 1; i < maxm; i++)
	{
		res += a[i] / 2;
		a[i] %= 2;

		mask += (1 << i) * a[i];
	}

	for(int i = 0; i < maxm; i++)
		And[1 << i] = i;

	for(int mask = 1; mask < (1 << maxm); mask++)
	{
		int lb = (mask&(-mask));

		if(lb == mask)
			continue;

		And[mask] = And[mask ^ lb] ^ And[lb];
	}

	for(int mask = 1; mask < (1 << maxm); mask++)
	{
		if(And[mask] != 0)
			continue;

		dp[mask] = 1;
		for(int submask = mask; submask > 0; submask = (submask - 1) & mask)
			if(submask != mask && And[submask] == 0)
				dp[mask] = max(dp[mask] , dp[submask] + 1);
	}

	cout << n - res - dp[mask] << endl;
}









```