#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 100005;
int n, x, y, a, b[MX], c[16], ans, d[1<<16], k, t;
int main(){
	scanf("%d", &n);
	rep(i,n-1){
		scanf("%d%d%d", &x, &y, &a);
		b[x] ^= a;
		b[y] ^= a;
	}
	rep(i,n) c[b[i]]++;
	for(int i=1; i<1<<16; i++){
		k = 0;
		rep(j,16) if(i>>j&1) k ^= j;
		if(!k){
			d[i] = 1;
			for(int j=(i-1)&i; j; j=(j-1)&i) d[i] = max(d[i], d[j]+d[j^i]);
		}
	}
	for(int i=1; i<16; i++){
		ans += c[i] / 2;
		if(c[i]&1) t ^= 1<<i;
	}
	printf("%d\n", ans + __builtin_popcount(t) - d[t]);
}
### 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;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 100005;
int n, x, y, a, b[MX], c[16], ans, d[1<<16], k, t;
int main(){
	scanf("%d", &n);
	rep(i,n-1){
		scanf("%d%d%d", &x, &y, &a);
		b[x] ^= a;
		b[y] ^= a;
	}
	rep(i,n) c[b[i]]++;
	for(int i=1; i<1<<16; i++){
		k = 0;
		rep(j,16) if(i>>j&1) k ^= j;
		if(!k){
			d[i] = 1;
			for(int j=(i-1)&i; j; j=(j-1)&i) d[i] = max(d[i], d[j]+d[j^i]);
		}
	}
	for(int i=1; i<16; i++){
		ans += c[i] / 2;
		if(c[i]&1) t ^= 1<<i;
	}
	printf("%d\n", ans + __builtin_popcount(t) - d[t]);
}
```