#include<bits/stdc++.h>
#define ll long long
using namespace std;
int inf,n,sz[101],f[200010],ans,hh,a[200010];
int dfs(int now){
	int i,j,x,nt;
	if(!now)return 0;
	if(f[now]!=inf)return f[now];
	for(i=0;i<16;i++)if(now&(1<<i))
	 for(j=0;j<16;j++)if(j!=i&&(now&(1<<j))){
	 	x=i^j;
	 	nt=now^(1<<i)^(1<<j)^(1<<x);
	 	if(now&(1<<x))f[now]=min(f[now],dfs(nt)+2);
	 	 else f[now]=min(f[now],dfs(nt)+1);
	 }
	 return f[now];
}
int main(){
	int i,x,y,w;
	scanf("%d",&n);
	for(i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&w);
		a[x]^=w;a[y]^=w;
	}
	for(i=0;i<n;i++)sz[a[i]]++;
	for(i=1;i<16;i++)ans+=sz[i]/2,hh+=(1<<i)*(sz[i]%2);
	memset(f,44,sizeof(f));
	inf=f[0];
	printf("%d",ans+dfs(hh));
}
### Prompt

Generate a cpp solution to 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>
#define ll long long
using namespace std;
int inf,n,sz[101],f[200010],ans,hh,a[200010];
int dfs(int now){
	int i,j,x,nt;
	if(!now)return 0;
	if(f[now]!=inf)return f[now];
	for(i=0;i<16;i++)if(now&(1<<i))
	 for(j=0;j<16;j++)if(j!=i&&(now&(1<<j))){
	 	x=i^j;
	 	nt=now^(1<<i)^(1<<j)^(1<<x);
	 	if(now&(1<<x))f[now]=min(f[now],dfs(nt)+2);
	 	 else f[now]=min(f[now],dfs(nt)+1);
	 }
	 return f[now];
}
int main(){
	int i,x,y,w;
	scanf("%d",&n);
	for(i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&w);
		a[x]^=w;a[y]^=w;
	}
	for(i=0;i<n;i++)sz[a[i]]++;
	for(i=1;i<16;i++)ans+=sz[i]/2,hh+=(1<<i)*(sz[i]%2);
	memset(f,44,sizeof(f));
	inf=f[0];
	printf("%d",ans+dfs(hh));
}
```