#include<stdio.h>
#include<algorithm>
using namespace std;
int lowbit(int x){return x&-x;}
int a[100010],c[20],v[32768],s[32768],f[32768];
int main(){
	int n,i,j,x,y,z,res,M;
	scanf("%d",&n);
	for(i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&z);
		a[x+1]^=z;
		a[y+1]^=z;
	}
	for(i=1;i<=n;i++)c[a[i]]++;
	res=M=0;
	res+=c[0];
	for(i=1;i<16;i++){
		res+=c[i]/2;
		if(c[i]&1)v[1<<(M++)]=i;
	}
	for(i=1;i<1<<M;i++)s[i]=s[i^lowbit(i)]^v[lowbit(i)];
	for(i=1;i<1<<M;i++){
		if(s[i]==0){
			for(j=i;j;j=(j-1)&i){
				if(s[j]==0)f[i]=max(f[i],f[i^j]+1);
			}
		}
	}
	printf("%d",n-(res+f[(1<<M)-1]));
}
### 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<stdio.h>
#include<algorithm>
using namespace std;
int lowbit(int x){return x&-x;}
int a[100010],c[20],v[32768],s[32768],f[32768];
int main(){
	int n,i,j,x,y,z,res,M;
	scanf("%d",&n);
	for(i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&z);
		a[x+1]^=z;
		a[y+1]^=z;
	}
	for(i=1;i<=n;i++)c[a[i]]++;
	res=M=0;
	res+=c[0];
	for(i=1;i<16;i++){
		res+=c[i]/2;
		if(c[i]&1)v[1<<(M++)]=i;
	}
	for(i=1;i<1<<M;i++)s[i]=s[i^lowbit(i)]^v[lowbit(i)];
	for(i=1;i<1<<M;i++){
		if(s[i]==0){
			for(j=i;j;j=(j-1)&i){
				if(s[j]==0)f[i]=max(f[i],f[i^j]+1);
			}
		}
	}
	printf("%d",n-(res+f[(1<<M)-1]));
}
```