#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=100005;

int dp[1<<16],V[MAXN],cnt[16];

int DP(int s)
{
	if(s==0)
		return 0;
	if(dp[s]!=-1)
		return dp[s];
	dp[s]=0x3FFFFFFF;
	for(int i=1;i<16;i++)
		if(s&(1<<i))
			for(int j=1;j<16;j++)
				if(s&(1<<j))
				{
					int ns=s^(1<<i)^(1<<j)^(1<<(i^j));
					if(__builtin_popcount(ns)<__builtin_popcount(s))
						dp[s]=min(dp[s],DP(ns)+1+(((1<<(i^j))&s)==(1<<(i^j))));
				}
	return dp[s];
}

int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1,x,y,a;i<n;i++)
	{
		scanf("%d%d%d",&x,&y,&a);
		V[x]^=a;
		V[y]^=a;
	}
	for(int i=0;i<n;i++)
		if(V[i])
			cnt[V[i]]++;
	int ans=0,st=0;
	for(int i=0;i<16;i++)
	{
		ans+=cnt[i]/2;
		st^=(cnt[i]&1)<<i;
	}
	memset(dp,-1,sizeof dp);
	ans+=DP(st);
	printf("%d\n",ans);
	
	return 0;
}

### Prompt

Please provide a CPP coded solution 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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=100005;

int dp[1<<16],V[MAXN],cnt[16];

int DP(int s)
{
	if(s==0)
		return 0;
	if(dp[s]!=-1)
		return dp[s];
	dp[s]=0x3FFFFFFF;
	for(int i=1;i<16;i++)
		if(s&(1<<i))
			for(int j=1;j<16;j++)
				if(s&(1<<j))
				{
					int ns=s^(1<<i)^(1<<j)^(1<<(i^j));
					if(__builtin_popcount(ns)<__builtin_popcount(s))
						dp[s]=min(dp[s],DP(ns)+1+(((1<<(i^j))&s)==(1<<(i^j))));
				}
	return dp[s];
}

int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1,x,y,a;i<n;i++)
	{
		scanf("%d%d%d",&x,&y,&a);
		V[x]^=a;
		V[y]^=a;
	}
	for(int i=0;i<n;i++)
		if(V[i])
			cnt[V[i]]++;
	int ans=0,st=0;
	for(int i=0;i<16;i++)
	{
		ans+=cnt[i]/2;
		st^=(cnt[i]&1)<<i;
	}
	memset(dp,-1,sizeof dp);
	ans+=DP(st);
	printf("%d\n",ans);
	
	return 0;
}

```