#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;

int n,ans;
int t[N],a[18];

int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		t[x]^=z;
		t[y]^=z;
	}
	for(int i=0;i<n;i++)
		a[t[i]]++;
	for(int i=1;i<=15;i++)
		ans+=a[i]>>1,a[i]&=1;
	for(int i=1;i<=15;i++)
	for(int j=i+1;j<=15;j++)
		if(a[i^j] && a[i] && a[j]) a[i]=0,a[j]=0,a[i^j]=0,ans+=2;
	for(int i=1;i<=15;i++)
	for(int j=i+1;j<=15;j++)
	for(int k=j+1;k<=15;k++)
		if(a[i] && a[j] && a[k] && a[i^j^k]) a[i]=0,a[j]=0,a[k]=0,a[i^j^k]=0,ans+=3;
	for(int i=1;i<=15;i++)
		if(a[i]) a[16]++;
	if(a[16]>=5) ans+=a[16]-1;
	else ans+=a[16];
	printf("%d\n",ans);
	
	return 0;
}
### 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>
using namespace std;
const int N=1e5+5;

int n,ans;
int t[N],a[18];

int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		t[x]^=z;
		t[y]^=z;
	}
	for(int i=0;i<n;i++)
		a[t[i]]++;
	for(int i=1;i<=15;i++)
		ans+=a[i]>>1,a[i]&=1;
	for(int i=1;i<=15;i++)
	for(int j=i+1;j<=15;j++)
		if(a[i^j] && a[i] && a[j]) a[i]=0,a[j]=0,a[i^j]=0,ans+=2;
	for(int i=1;i<=15;i++)
	for(int j=i+1;j<=15;j++)
	for(int k=j+1;k<=15;k++)
		if(a[i] && a[j] && a[k] && a[i^j^k]) a[i]=0,a[j]=0,a[k]=0,a[i^j^k]=0,ans+=3;
	for(int i=1;i<=15;i++)
		if(a[i]) a[16]++;
	if(a[16]>=5) ans+=a[16]-1;
	else ans+=a[16];
	printf("%d\n",ans);
	
	return 0;
}
```