#include<bits/stdc++.h>
using namespace std;

int N,ans;
int val[100005];
int S;
int A[100005],siz[100005],F[100005];

int main(){
	scanf("%d",&N);
	for(int i=1;i<N;i++){
		int u,v,c;scanf("%d%d%d",&u,&v,&c);
		val[u]^=c;val[v]^=c;
	}
	for(int i=0;i<N;i++)if(val[i]){S^=(1<<val[i]);if(!((S>>val[i])&1)) ans++;}
	for(int i=1;i<(1<<16);i++){
		int j=0;
		for(;;j++) if((i>>j)&1) break;
		A[i]=A[i^(1<<j)]^j; 
		siz[i]=siz[i>>1]+(i&1);
	}
	for(int i=1;i<(1<<16);i++)if(A[i]==0){
		F[i]=siz[i]-1;
		for(int j=(i-1)&i;j;j=(j-1)&i) if(A[j]==0)
			F[i]=min(F[i],F[j]+F[i^j]);
	}
	printf("%d\n",ans+F[S]);
}
### Prompt

Develop a solution in Cpp to the problem described below:

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;

int N,ans;
int val[100005];
int S;
int A[100005],siz[100005],F[100005];

int main(){
	scanf("%d",&N);
	for(int i=1;i<N;i++){
		int u,v,c;scanf("%d%d%d",&u,&v,&c);
		val[u]^=c;val[v]^=c;
	}
	for(int i=0;i<N;i++)if(val[i]){S^=(1<<val[i]);if(!((S>>val[i])&1)) ans++;}
	for(int i=1;i<(1<<16);i++){
		int j=0;
		for(;;j++) if((i>>j)&1) break;
		A[i]=A[i^(1<<j)]^j; 
		siz[i]=siz[i>>1]+(i&1);
	}
	for(int i=1;i<(1<<16);i++)if(A[i]==0){
		F[i]=siz[i]-1;
		for(int j=(i-1)&i;j;j=(j-1)&i) if(A[j]==0)
			F[i]=min(F[i],F[j]+F[i^j]);
	}
	printf("%d\n",ans+F[S]);
}
```