#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>

const int MAXN = 100000;
const int MAXA = 15;
const int INF = 0x3f3f3f3f;

int N;
int A[MAXN + 5];
int Cnt[MAXA + 5];

int Dp[(1 << (MAXA + 1)) + 5];

int Dfs(int S) {
	if (Dp[S] != INF)
		return Dp[S];
	for (int i = 1; i <= MAXA; i++)
		if ((S >> i) & 1)
			for (int j = i + 1; j <= MAXA; j++)
				if (((S >> j) & 1))
					Dp[S] = std::min(Dp[S], Dfs(S ^ (1 << i) ^ (1 << j) ^ (1 << (i ^ j))) + 1 + ((S >> (i ^ j)) & 1));
	return Dp[S];
}

int main() {
	scanf("%d", &N);
	for (int i = 1; i < N; i++) {
		int u, v, w; scanf("%d%d%d", &u, &v, &w);
		A[u + 1] ^= w, A[v + 1] ^= w;
	}
	for (int i = 1; i <= N; i++)
		Cnt[A[i]]++;
	int Ans = 0, S = 0;
	for (int i = 1; i <= MAXA; i++) {
		Ans += Cnt[i] / 2;
		Cnt[i] %= 2;
		if (Cnt[i])
			S |= (1 << i);
	}
	memset(Dp, 0x3f, sizeof Dp);
	Dp[0] = 0;
	printf("%d", Dfs(S) + Ans);
	return 0;
}

### Prompt

Generate 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 <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>

const int MAXN = 100000;
const int MAXA = 15;
const int INF = 0x3f3f3f3f;

int N;
int A[MAXN + 5];
int Cnt[MAXA + 5];

int Dp[(1 << (MAXA + 1)) + 5];

int Dfs(int S) {
	if (Dp[S] != INF)
		return Dp[S];
	for (int i = 1; i <= MAXA; i++)
		if ((S >> i) & 1)
			for (int j = i + 1; j <= MAXA; j++)
				if (((S >> j) & 1))
					Dp[S] = std::min(Dp[S], Dfs(S ^ (1 << i) ^ (1 << j) ^ (1 << (i ^ j))) + 1 + ((S >> (i ^ j)) & 1));
	return Dp[S];
}

int main() {
	scanf("%d", &N);
	for (int i = 1; i < N; i++) {
		int u, v, w; scanf("%d%d%d", &u, &v, &w);
		A[u + 1] ^= w, A[v + 1] ^= w;
	}
	for (int i = 1; i <= N; i++)
		Cnt[A[i]]++;
	int Ans = 0, S = 0;
	for (int i = 1; i <= MAXA; i++) {
		Ans += Cnt[i] / 2;
		Cnt[i] %= 2;
		if (Cnt[i])
			S |= (1 << i);
	}
	memset(Dp, 0x3f, sizeof Dp);
	Dp[0] = 0;
	printf("%d", Dfs(S) + Ans);
	return 0;
}

```