#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define N_ 101000
#define pii pair<int,int>
int w[N_], C[N_], D[1<<16], P[N_], cnt, CC[N_];
int n;
priority_queue<pii>PQ;
void Put(int mask, int d) {
	if (D[mask] <= d)return;
	D[mask] = d;
	PQ.push(pii(-d, mask));
}
int main() {
	int i, a, b, c, j;
	scanf("%d", &n);
	for (i = 1; i < n; i++) {
		scanf("%d%d%d", &a, &b, &c);
		a++, b++;
		w[a] ^= c;
		w[b] ^= c;
	}
	int s = 0, res = 0;
	for (i = 1; i <= n; i++) {
		if (w[i])C[w[i]]++, res++;
	}
	for (i = 1; i <= 15; i++)if (C[i] & 1)s |= (1 << i);
	for (i = 0; i < (1 << 16); i++) {
		D[i] = 1e9;
		for (j = 0; j < 16; j++)if ((i >> j) & 1)CC[i]++;
	}
	Put(s, 0);
	for (i = 2; i < (1 << 16); i += 2) {
		int t = 0;
		for (j = 1; j <= 15; j++)if ((i >> j) & 1)t ^= j;
		if (t == 0)P[cnt++] = i;
	}
	while (!PQ.empty()) {
		pii tp = PQ.top();
		PQ.pop();
		int x = tp.second;
		if (-tp.first != D[x])continue;
		for (i = 0; i < cnt; i++) {
			int y = P[i];
			if (!(x&y))continue;
			Put(y^x, D[x] + CC[y] - 2);
		}
	}
	printf("%d\n", (res + D[0])/2);
	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<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define N_ 101000
#define pii pair<int,int>
int w[N_], C[N_], D[1<<16], P[N_], cnt, CC[N_];
int n;
priority_queue<pii>PQ;
void Put(int mask, int d) {
	if (D[mask] <= d)return;
	D[mask] = d;
	PQ.push(pii(-d, mask));
}
int main() {
	int i, a, b, c, j;
	scanf("%d", &n);
	for (i = 1; i < n; i++) {
		scanf("%d%d%d", &a, &b, &c);
		a++, b++;
		w[a] ^= c;
		w[b] ^= c;
	}
	int s = 0, res = 0;
	for (i = 1; i <= n; i++) {
		if (w[i])C[w[i]]++, res++;
	}
	for (i = 1; i <= 15; i++)if (C[i] & 1)s |= (1 << i);
	for (i = 0; i < (1 << 16); i++) {
		D[i] = 1e9;
		for (j = 0; j < 16; j++)if ((i >> j) & 1)CC[i]++;
	}
	Put(s, 0);
	for (i = 2; i < (1 << 16); i += 2) {
		int t = 0;
		for (j = 1; j <= 15; j++)if ((i >> j) & 1)t ^= j;
		if (t == 0)P[cnt++] = i;
	}
	while (!PQ.empty()) {
		pii tp = PQ.top();
		PQ.pop();
		int x = tp.second;
		if (-tp.first != D[x])continue;
		for (i = 0; i < cnt; i++) {
			int y = P[i];
			if (!(x&y))continue;
			Put(y^x, D[x] + CC[y] - 2);
		}
	}
	printf("%d\n", (res + D[0])/2);
	return 0;
}
```