#include <bits/stdc++.h>
#define ll long long
using namespace std;
int read()
{
	int x=0,f=1;
	char c=getchar();
	for(;c<'0'||c>'9';c=getchar())if(c=='-')f=-1;
	for(;c>='0'&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
const int N=2e5;
int ans,n,a[N],t,f[1<<20],q[N],d[N];
int main()
{
	n=read();
	int x,y,v;
	for(int i=1;i<n;++i)x=read()+1,y=read()+1,v=read(),a[x]^=v,a[y]^=v;
	sort(a+1,a+n+1);//for(int i=1;i<=n;++i)printf("%d ",a[i]);printf("\n");
	for(int i=1;i<=n;++i)
	if(i<n&&a[i]==a[i+1]) a[i]!=0?ans++:0,i++;
	else q[++t]=a[i];
	for(int i=2;i<(1<<t);++i)f[i]=f[i>>1]+(i&1);
	for(int i=1;i<(1<<t);++i)
	for(int j=1;j<=t;++j)
	if(i&(1<<(j-1)))
	d[i]^=q[j];
	for(int i=1;i<(1<<t);++i)
	if(!d[i])
	for(int j=(i-1)&i;j;j=(j-1)&i)
	if(!d[j])
	f[i]=min(f[i],f[j]+f[i^j]);
	printf("%d\n",f[(1<<t)-1]+ans);
}
### Prompt

Your challenge is to write 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 read()
{
	int x=0,f=1;
	char c=getchar();
	for(;c<'0'||c>'9';c=getchar())if(c=='-')f=-1;
	for(;c>='0'&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
const int N=2e5;
int ans,n,a[N],t,f[1<<20],q[N],d[N];
int main()
{
	n=read();
	int x,y,v;
	for(int i=1;i<n;++i)x=read()+1,y=read()+1,v=read(),a[x]^=v,a[y]^=v;
	sort(a+1,a+n+1);//for(int i=1;i<=n;++i)printf("%d ",a[i]);printf("\n");
	for(int i=1;i<=n;++i)
	if(i<n&&a[i]==a[i+1]) a[i]!=0?ans++:0,i++;
	else q[++t]=a[i];
	for(int i=2;i<(1<<t);++i)f[i]=f[i>>1]+(i&1);
	for(int i=1;i<(1<<t);++i)
	for(int j=1;j<=t;++j)
	if(i&(1<<(j-1)))
	d[i]^=q[j];
	for(int i=1;i<(1<<t);++i)
	if(!d[i])
	for(int j=(i-1)&i;j;j=(j-1)&i)
	if(!d[j])
	f[i]=min(f[i],f[j]+f[i^j]);
	printf("%d\n",f[(1<<t)-1]+ans);
}
```