#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;
const int Maxn = 100010;
const int inf = 0x7fffffff;
int po[Maxn];
int f[1<<15], bit[1<<15], n;
vector <int> vec[20];
void down(int &x, int y) { if(x > y) x = y; }
int main() {
	int i, j, k;
	scanf("%d", &n);
	for(i = 1; i < n; i++){
		int x, y, d;
		scanf("%d%d%d", &x, &y, &d);
		po[x] ^= d; po[y] ^= d;
	}
	for(i = 1; i < 1<<15; i++){ bit[i] = bit[i>>1]+(i&1); vec[bit[i]].push_back(i); }
	int p = 0, ans = 0;
	for(i = 0; i < 1<<15; i++) f[i] = inf;
	for(i = 0; i < n; i++){
		if(!po[i]) continue;
		if(p&(1<<po[i]-1)) ans++;
		p ^= (1<<po[i]-1);
	}
	f[p] = 0;
	for(i = 15; i >= 1; i--){
		for(j = 0; j < vec[i].size(); j++){
			int x = vec[i][j];
			if(f[x] == inf) continue;
			for(k = 0; k < 15; k++){
				if(x&(1<<k)){
					for(int u = 0; u < 15; u++){
						if(k != u && x&(1<<u)){
							int t = (k+1)^(u+1);
							if(x&(1<<t-1)) down(f[x^(1<<k)^(1<<u)^(1<<(t-1))], f[x]+2);
							else down(f[x^(1<<k)^(1<<u)^(1<<(t-1))], f[x]+1);
						}
					}
				}
			}
		}
	}
	printf("%d\n", f[0]+ans);
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;
const int Maxn = 100010;
const int inf = 0x7fffffff;
int po[Maxn];
int f[1<<15], bit[1<<15], n;
vector <int> vec[20];
void down(int &x, int y) { if(x > y) x = y; }
int main() {
	int i, j, k;
	scanf("%d", &n);
	for(i = 1; i < n; i++){
		int x, y, d;
		scanf("%d%d%d", &x, &y, &d);
		po[x] ^= d; po[y] ^= d;
	}
	for(i = 1; i < 1<<15; i++){ bit[i] = bit[i>>1]+(i&1); vec[bit[i]].push_back(i); }
	int p = 0, ans = 0;
	for(i = 0; i < 1<<15; i++) f[i] = inf;
	for(i = 0; i < n; i++){
		if(!po[i]) continue;
		if(p&(1<<po[i]-1)) ans++;
		p ^= (1<<po[i]-1);
	}
	f[p] = 0;
	for(i = 15; i >= 1; i--){
		for(j = 0; j < vec[i].size(); j++){
			int x = vec[i][j];
			if(f[x] == inf) continue;
			for(k = 0; k < 15; k++){
				if(x&(1<<k)){
					for(int u = 0; u < 15; u++){
						if(k != u && x&(1<<u)){
							int t = (k+1)^(u+1);
							if(x&(1<<t-1)) down(f[x^(1<<k)^(1<<u)^(1<<(t-1))], f[x]+2);
							else down(f[x^(1<<k)^(1<<u)^(1<<(t-1))], f[x]+1);
						}
					}
				}
			}
		}
	}
	printf("%d\n", f[0]+ans);
	return 0;
}
```