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

const int N = 1e5 + 100, M = 16;

int n, ans, a[N], cnt[M], dp[1 << M];

bool ok(int mask) {
	int res = 0;
	for (int i = 0; i < M; i++)
		if (mask & (1 << i))
			res ^= i;
	return res == 0;
}

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int u, v, w;
		cin >> u >> v >> w;
		a[u] ^= w;
		a[v] ^= w;
	}
	for (int i = 0; i < n; i++)
		cnt[a[i]]++;
	for (int i = 1; i < M; i++) {
		ans += (cnt[i] >> 1);
		cnt[i] &= 1;
	}
	int mask = 0;
	for (int i = 1; i < M; i++)
		if (cnt[i])
			mask |= 1 << i;
	ans += __builtin_popcount(mask);
	for (int x = 1; x <= mask; x++) {
		dp[x] = ok(x);
		if ((mask & x) == x)
			for (int y = x; y; y = (--y & x))
				if (y < x && ok(y))
					dp[x] = max(dp[x], dp[y] + dp[x ^ y]);
	}
	ans -= dp[mask];
	cout << ans;
}

### Prompt

Develop a solution in cpp 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 <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 100, M = 16;

int n, ans, a[N], cnt[M], dp[1 << M];

bool ok(int mask) {
	int res = 0;
	for (int i = 0; i < M; i++)
		if (mask & (1 << i))
			res ^= i;
	return res == 0;
}

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int u, v, w;
		cin >> u >> v >> w;
		a[u] ^= w;
		a[v] ^= w;
	}
	for (int i = 0; i < n; i++)
		cnt[a[i]]++;
	for (int i = 1; i < M; i++) {
		ans += (cnt[i] >> 1);
		cnt[i] &= 1;
	}
	int mask = 0;
	for (int i = 1; i < M; i++)
		if (cnt[i])
			mask |= 1 << i;
	ans += __builtin_popcount(mask);
	for (int x = 1; x <= mask; x++) {
		dp[x] = ok(x);
		if ((mask & x) == x)
			for (int y = x; y; y = (--y & x))
				if (y < x && ok(y))
					dp[x] = max(dp[x], dp[y] + dp[x ^ y]);
	}
	ans -= dp[mask];
	cout << ans;
}

```