#include<bits/stdc++.h>
#define N 100005
using namespace std;
int a[N],c[N];
int n,m,v[20];
int f[N],ans;
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		a[x]^=z; a[y]^=z;
	}
	for (int i=0;i<n;i++)
		c[a[i]]++;
	for (int i=1;i<16;i++)
	{
		ans+=c[i]/2;
		if (c[i]&1) v[m++]=i;
	}
	f[0]=0;
	for (int i=1;i<1<<m;i++)
	{
		int k=0,x=0;
		for (int j=0;j<m;j++)
			if ((i>>j)&1)
			{
				x^=v[j];
				k++;
			}
		f[i]=(x?100:k-1);
		for (int p=i-(i&(-i)),j=p;;j=(j-1)&p)
		{
			f[i]=min(f[i],f[j]+f[i^j]);
			if (!j) break; 
		}
	}
	printf("%d",ans+f[(1<<m)-1]);
}
### Prompt

Your task is to create 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 N 100005
using namespace std;
int a[N],c[N];
int n,m,v[20];
int f[N],ans;
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		a[x]^=z; a[y]^=z;
	}
	for (int i=0;i<n;i++)
		c[a[i]]++;
	for (int i=1;i<16;i++)
	{
		ans+=c[i]/2;
		if (c[i]&1) v[m++]=i;
	}
	f[0]=0;
	for (int i=1;i<1<<m;i++)
	{
		int k=0,x=0;
		for (int j=0;j<m;j++)
			if ((i>>j)&1)
			{
				x^=v[j];
				k++;
			}
		f[i]=(x?100:k-1);
		for (int p=i-(i&(-i)),j=p;;j=(j-1)&p)
		{
			f[i]=min(f[i],f[j]+f[i^j]);
			if (!j) break; 
		}
	}
	printf("%d",ans+f[(1<<m)-1]);
}
```