#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,b[1<<16],f[1<<16],s;
int a[100001],o[16],ans,num[1<<16],inf;
inline bool cmp(int x,int y){return num[x]<num[y];}
int dp(int i){
	if (f[i]<inf) return f[i]; 
	for (int x=1; x<=15; x++){
		if (((1<<x)&i)==0) continue;
		for (int y=1; y<=15; y++){
			if (x==y) continue;
			if (((1<<y)&i)==0) continue;
			int z=x^y;
			int p=(((i^(1<<z))^(1<<x))^(1<<y));
			if ((1<<z)&i)
				f[i]=min(f[i],dp(p)+2);
			else
				f[i]=min(f[i],dp(p)+1);
		}
	}
	return f[i];
}
int main(){
	scanf("%d",&n);
	for (int x,y,z,i=1; i<n; i++){
		scanf("%d%d%d",&x,&y,&z);
		x++,y++;
		a[x]^=z,a[y]^=z;
	}
	for (int i=1; i<=n; i++) o[a[i]]++;
	for (int i=1; i<=15; i++){
		if (o[i]>=2){
			ans+=o[i]/2; o[i]=o[i]&1;
		}
		if (o[i]) s|=(1<<i);
	}
	memset(f,10,sizeof(f));
	int ful=(1<<16)-1; f[0]=0; inf=f[1];
	printf("%d\n",ans+dp(s));
	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<iostream>
#include<algorithm>
using namespace std;
int n,b[1<<16],f[1<<16],s;
int a[100001],o[16],ans,num[1<<16],inf;
inline bool cmp(int x,int y){return num[x]<num[y];}
int dp(int i){
	if (f[i]<inf) return f[i]; 
	for (int x=1; x<=15; x++){
		if (((1<<x)&i)==0) continue;
		for (int y=1; y<=15; y++){
			if (x==y) continue;
			if (((1<<y)&i)==0) continue;
			int z=x^y;
			int p=(((i^(1<<z))^(1<<x))^(1<<y));
			if ((1<<z)&i)
				f[i]=min(f[i],dp(p)+2);
			else
				f[i]=min(f[i],dp(p)+1);
		}
	}
	return f[i];
}
int main(){
	scanf("%d",&n);
	for (int x,y,z,i=1; i<n; i++){
		scanf("%d%d%d",&x,&y,&z);
		x++,y++;
		a[x]^=z,a[y]^=z;
	}
	for (int i=1; i<=n; i++) o[a[i]]++;
	for (int i=1; i<=15; i++){
		if (o[i]>=2){
			ans+=o[i]/2; o[i]=o[i]&1;
		}
		if (o[i]) s|=(1<<i);
	}
	memset(f,10,sizeof(f));
	int ful=(1<<16)-1; f[0]=0; inf=f[1];
	printf("%d\n",ans+dp(s));
	return 0;
}
```