#include <bits/stdc++.h>
#define N 100010
using namespace std;

inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
int d[N],n,ct[20],Ans,F[N],S[N];
int main()
{
	// freopen("read.in","r",stdin);
	for(int i=1;i<(1<<16);i++)
		F[i]=F[i-(i&-i)]^(__builtin_popcount((i&-i)-1));
	// for(int i=0;i<=10;i++) printf("%d\n",F[i]);
	n=read(); for(int i=1;i<n;i++)
	{
		int u=read()+1, v=read()+1, w=read();
		d[u]^=w; d[v]^=w;
	}
	// for(int i=1;i<=n;i++) printf("%d ",d[i]); puts("");
	for(int i=1;i<=n;i++) ct[d[i]]++;
	// for(int i=1;i<16;i++) printf("%d ",ct[i]); puts("");
	for(int i=1;i<16;i++) Ans+=ct[i]/2,ct[i]&=1;
	// memset(S,0x3f,sizeof S); S[0]=0;
	for(int i=1;i<(1<<16);i++)
	{
		for(int j=i;j;j=(j-1)&i)
			if(!F[j]) S[i]=max(S[i],S[i^j]+1);
	}
	int Ret=0; for(int i=1;i<16;i++) Ret|=(ct[i]<<i);
	// cout << Ret << endl;
	cout << Ans+__builtin_popcount(Ret)-S[Ret] << endl;
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 100010
using namespace std;

inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
int d[N],n,ct[20],Ans,F[N],S[N];
int main()
{
	// freopen("read.in","r",stdin);
	for(int i=1;i<(1<<16);i++)
		F[i]=F[i-(i&-i)]^(__builtin_popcount((i&-i)-1));
	// for(int i=0;i<=10;i++) printf("%d\n",F[i]);
	n=read(); for(int i=1;i<n;i++)
	{
		int u=read()+1, v=read()+1, w=read();
		d[u]^=w; d[v]^=w;
	}
	// for(int i=1;i<=n;i++) printf("%d ",d[i]); puts("");
	for(int i=1;i<=n;i++) ct[d[i]]++;
	// for(int i=1;i<16;i++) printf("%d ",ct[i]); puts("");
	for(int i=1;i<16;i++) Ans+=ct[i]/2,ct[i]&=1;
	// memset(S,0x3f,sizeof S); S[0]=0;
	for(int i=1;i<(1<<16);i++)
	{
		for(int j=i;j;j=(j-1)&i)
			if(!F[j]) S[i]=max(S[i],S[i^j]+1);
	}
	int Ret=0; for(int i=1;i<16;i++) Ret|=(ct[i]<<i);
	// cout << Ret << endl;
	cout << Ans+__builtin_popcount(Ret)-S[Ret] << endl;
	return 0;
}
```