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

const int MAXN = 101010;

int a[MAXN];
int cnt[21];
int f[1<<17];
int n;

int res[21];
int cntr;

int main() {
	scanf("%d", &n);
	int u, v, w;
	for(int i = 1; i < n; ++i) {
		scanf("%d%d%d", &u, &v, &w);
		a[u] ^= w;
		a[v] ^= w;
	}
	for(int i = 0; i < n; ++i)
		cnt[a[i]]++;
	int ans = 0;
	for(int i = 1; i <= 15; ++i) {
		ans += cnt[i] >> 1;
		if(cnt[i] & 1) {
			res[cntr++] = i;
		}
	}
	f[0] = 0;
	for(int i = 1; i < 1<<cntr; ++i) {
		int tmp = 0;
		for(int j = 0; j < cntr; ++j) {
			if(i & (1<<j))
				tmp ^= res[j];
		}
		tmp = !tmp;
		for(int j = 0; j < cntr; ++j) {
			if(i & (1<<j)) {
				f[i] = max(f[i], f[i ^ (1<<j)] + tmp);
			}
		}
	}
	ans += cntr;
	ans -= f[(1<<cntr) - 1];
	printf("%d\n", ans);
	
	return 0;
}
### Prompt

Please formulate 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;

const int MAXN = 101010;

int a[MAXN];
int cnt[21];
int f[1<<17];
int n;

int res[21];
int cntr;

int main() {
	scanf("%d", &n);
	int u, v, w;
	for(int i = 1; i < n; ++i) {
		scanf("%d%d%d", &u, &v, &w);
		a[u] ^= w;
		a[v] ^= w;
	}
	for(int i = 0; i < n; ++i)
		cnt[a[i]]++;
	int ans = 0;
	for(int i = 1; i <= 15; ++i) {
		ans += cnt[i] >> 1;
		if(cnt[i] & 1) {
			res[cntr++] = i;
		}
	}
	f[0] = 0;
	for(int i = 1; i < 1<<cntr; ++i) {
		int tmp = 0;
		for(int j = 0; j < cntr; ++j) {
			if(i & (1<<j))
				tmp ^= res[j];
		}
		tmp = !tmp;
		for(int j = 0; j < cntr; ++j) {
			if(i & (1<<j)) {
				f[i] = max(f[i], f[i ^ (1<<j)] + tmp);
			}
		}
	}
	ans += cntr;
	ans -= f[(1<<cntr) - 1];
	printf("%d\n", ans);
	
	return 0;
}
```