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

int n, x, y, z, ans, f[1 << 15], a[100100], c[20];
vector<int> o;

int main() {
	scanf("%d", &n);
	for(int i = 1; i < n; i++) {
		scanf("%d%d%d", &x, &y, &z);
		a[x] ^= z, a[y] ^= z;
	}
	for(int i = 0; i < n; i++)
		c[a[i]]++;
	for(int i = 1; i <= 15; i++) {
		ans += c[i] / 2;
//		printf("%d %d\n", i, c[i]);
		if(c[i] & 1) o.push_back(i);
	}
	f[0] = 0;
	for(int i = 1; i < (1 << o.size()); i++) {
		int x = 0, k = 0;
		for(int j = 0; j < o.size(); j++)
			if(i & (1 << j)) x ^= o[j], k++;
		f[i] = x ? 19260817 : k - 1;
//		if(i == 7) printf("%d\n", (1 << o.size()) - 1);
		for(int j = i - (i & -i); j; j = (i - (i & -i)) & (j - 1))
			f[i] = min(f[i], f[j] + f[j ^ i]);
	}
	printf("%d\n", f[(1 << o.size()) - 1] + ans);
	return 0;
}
### Prompt

In cpp, your task is to solve 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, x, y, z, ans, f[1 << 15], a[100100], c[20];
vector<int> o;

int main() {
	scanf("%d", &n);
	for(int i = 1; i < n; i++) {
		scanf("%d%d%d", &x, &y, &z);
		a[x] ^= z, a[y] ^= z;
	}
	for(int i = 0; i < n; i++)
		c[a[i]]++;
	for(int i = 1; i <= 15; i++) {
		ans += c[i] / 2;
//		printf("%d %d\n", i, c[i]);
		if(c[i] & 1) o.push_back(i);
	}
	f[0] = 0;
	for(int i = 1; i < (1 << o.size()); i++) {
		int x = 0, k = 0;
		for(int j = 0; j < o.size(); j++)
			if(i & (1 << j)) x ^= o[j], k++;
		f[i] = x ? 19260817 : k - 1;
//		if(i == 7) printf("%d\n", (1 << o.size()) - 1);
		for(int j = i - (i & -i); j; j = (i - (i & -i)) & (j - 1))
			f[i] = min(f[i], f[j] + f[j ^ i]);
	}
	printf("%d\n", f[(1 << o.size()) - 1] + ans);
	return 0;
}
```