#include<bits/stdc++.h>
using namespace std;
#define N 100005
int va[N];
int num[20];
int a[20],sz,b[1001000];
int dp[1001000];
int solve(int s)
{
	if(s==0||dp[s]) return dp[s];
	dp[s]=1;
	for(int i=s;i;)
	{
		i=(i-1)&s;
        if(b[i]==0) dp[s]=max(dp[s],solve(i)+solve(s^i));
	}
	return dp[s];
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<n;++i)
	{
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		va[u]^=w; va[v]^=w;
	}
	for(int i=0;i<n;++i) ++num[va[i]];
	int cnt=num[0];
	for(int i=1;i<=15;++i)
	{
		cnt+=num[i]/2;
		if(num[i]&1) a[sz++]=i;
	}
	for(int s=0;s<(1<<sz);++s)
		for(int i=0;i<sz;++i)
			if(s>>i&1)
				b[s]^=a[i];
	if(sz)cnt+=solve((1<<sz)-1);
	printf("%d\n",n-cnt);
	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<bits/stdc++.h>
using namespace std;
#define N 100005
int va[N];
int num[20];
int a[20],sz,b[1001000];
int dp[1001000];
int solve(int s)
{
	if(s==0||dp[s]) return dp[s];
	dp[s]=1;
	for(int i=s;i;)
	{
		i=(i-1)&s;
        if(b[i]==0) dp[s]=max(dp[s],solve(i)+solve(s^i));
	}
	return dp[s];
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<n;++i)
	{
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		va[u]^=w; va[v]^=w;
	}
	for(int i=0;i<n;++i) ++num[va[i]];
	int cnt=num[0];
	for(int i=1;i<=15;++i)
	{
		cnt+=num[i]/2;
		if(num[i]&1) a[sz++]=i;
	}
	for(int s=0;s<(1<<sz);++s)
		for(int i=0;i<sz;++i)
			if(s>>i&1)
				b[s]^=a[i];
	if(sz)cnt+=solve((1<<sz)-1);
	printf("%d\n",n-cnt);
	return 0;
}

```