#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long LL;
const int N = 100050;

int a[N], n, s[17], f[1 << 15], siz[1 << 15];
int XOR_S[1 << 15], ans, zt, rev[N];
int main()
{

	int i, j, k, h;
	scanf("%d", &n);
	for(i = 1; i < n; i ++){
		scanf("%d%d%d", &j, &k, &h);
		j ++, k ++, a[j] ^= h, a[k] ^= h;
	}
	for(i = 1; i <= n; i ++)
		s[a[i]] ++;
	for(i = 1; i <= 15; i ++){
		ans += s[i] >> 1, rev[1 << i - 1] = i;
		zt |= (s[i] & 1) << i - 1;
	}
	for(i = 1; i < 1 << 15; i ++){
		XOR_S[i] = XOR_S[i - (i & -i)] ^ rev[i & -i];
		siz[i] = siz[i - (i & -i)] + 1;
	}
	for(i = 1; i <= zt; i ++){
		f[i] = 1000000000;
		for(j = i; j; j = (j - 1) & i)
			if(XOR_S[j] == 0)
				f[i] = min(f[i], f[i - j] + siz[j] - 1);
	}
	printf("%d", f[zt] + ans);

	return 0;
}

### Prompt

Please provide a Cpp coded solution 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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long LL;
const int N = 100050;

int a[N], n, s[17], f[1 << 15], siz[1 << 15];
int XOR_S[1 << 15], ans, zt, rev[N];
int main()
{

	int i, j, k, h;
	scanf("%d", &n);
	for(i = 1; i < n; i ++){
		scanf("%d%d%d", &j, &k, &h);
		j ++, k ++, a[j] ^= h, a[k] ^= h;
	}
	for(i = 1; i <= n; i ++)
		s[a[i]] ++;
	for(i = 1; i <= 15; i ++){
		ans += s[i] >> 1, rev[1 << i - 1] = i;
		zt |= (s[i] & 1) << i - 1;
	}
	for(i = 1; i < 1 << 15; i ++){
		XOR_S[i] = XOR_S[i - (i & -i)] ^ rev[i & -i];
		siz[i] = siz[i - (i & -i)] + 1;
	}
	for(i = 1; i <= zt; i ++){
		f[i] = 1000000000;
		for(j = i; j; j = (j - 1) & i)
			if(XOR_S[j] == 0)
				f[i] = min(f[i], f[i - j] + siz[j] - 1);
	}
	printf("%d", f[zt] + ans);

	return 0;
}

```