#include <bits/stdc++.h>
using namespace std;
vector<int> a[100020];
int n, x, y, z;
int w[100020];
int c[16];
int f[65537];
int g[65537];
int ans, mask;
int dfs(int x, int y) {
	int re = w[x];
	for (int i: a[x]) {
		if (i != y) {
			re ^= dfs(i, x);
		}
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		scanf("%d%d%d", &x, &y, &z);
		a[x].push_back(y);
		a[y].push_back(x);
		w[x] ^= z;
		w[y] ^= z;
	}
	for (int i = 0; i < n; i++) {
		c[w[i]]++;
	}
	for (int i = 1; i < 16; i++) {
		ans += c[i] / 2;
		c[i] %= 2;
		mask |= c[i] << i;
	}
	for (int i = 1; i < 1 << 16; i++) {
		int v = 0, c = 0;
		for (int j = 0; j < 16; j++) {
			if (i >> j & 1) {
				v ^= j;
				c++;
			}
		}
		if (v == 0) {
			f[i] = c - 1;
		} else {
			f[i] = -1;
		}
		for (int j = i; ; j = (j - 1) & i) {
			if (f[j] >= 0 && f[i ^ j] >= 0) {
				f[i] = min(f[i], f[j] + f[i ^ j]);
			}
			if (j == 0) {
				break;
			}
		}
	}
	printf("%d\n", ans + f[mask]);
//	dfs(0, -1);
	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;
vector<int> a[100020];
int n, x, y, z;
int w[100020];
int c[16];
int f[65537];
int g[65537];
int ans, mask;
int dfs(int x, int y) {
	int re = w[x];
	for (int i: a[x]) {
		if (i != y) {
			re ^= dfs(i, x);
		}
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		scanf("%d%d%d", &x, &y, &z);
		a[x].push_back(y);
		a[y].push_back(x);
		w[x] ^= z;
		w[y] ^= z;
	}
	for (int i = 0; i < n; i++) {
		c[w[i]]++;
	}
	for (int i = 1; i < 16; i++) {
		ans += c[i] / 2;
		c[i] %= 2;
		mask |= c[i] << i;
	}
	for (int i = 1; i < 1 << 16; i++) {
		int v = 0, c = 0;
		for (int j = 0; j < 16; j++) {
			if (i >> j & 1) {
				v ^= j;
				c++;
			}
		}
		if (v == 0) {
			f[i] = c - 1;
		} else {
			f[i] = -1;
		}
		for (int j = i; ; j = (j - 1) & i) {
			if (f[j] >= 0 && f[i ^ j] >= 0) {
				f[i] = min(f[i], f[j] + f[i ^ j]);
			}
			if (j == 0) {
				break;
			}
		}
	}
	printf("%d\n", ans + f[mask]);
//	dfs(0, -1);
	return 0;
}
```