#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int c[101000],ok[10000000];
int h[101000],nxt[201000],K=0,to[201000],cost[201000];int cnt[16],ans=0;
void ins(int u,int v,int c){nxt[++K]=h[u];h[u]=K;to[K]=v;cost[K]=c;}
void dfs(int x,int f,int val)
{
	int k=0;
	for(int i=h[x];i;i=nxt[i])
	{
		if(to[i]==f)continue;
		k^=cost[i];dfs(to[i],x,cost[i]);
	}
	if(x)cnt[val^k]++;
}
int a[18],tot=0,dp[1<<15];
int DP(int S)
{
	
	if(dp[S]!=-1)return dp[S];dp[S]=2000000000;
	for(int i=0;i<15;i++)
	for(int j=i+1;j<15;j++)
	{
		if(((S>>i)&1)&&((S>>j)&1))
		{
			int k=(i+1)^(j+1);
			dp[S]=min(dp[S],DP(S^(1<<(k-1))^(1<<j)^(1<<i))+1+(((S>>(k-1))&1)?1:0));
		}
	}
	for(int i=0;i<15;i++)
		if((S>>i)&1)
			dp[S]=min(dp[S],DP(S^(1<<i))+1);
	return dp[S];
}
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<n;i++){int x,y,c;scanf("%d%d%d",&x,&y,&c);ins(x,y,c);ins(y,x,c);}
	dfs(0,-1,0);
	int S=0;
	for(int i=1;i<=15;i++)
	{
		ans+=cnt[i]/2;
		if(cnt[i]&1)S|=1<<(i-1);
	}
	memset(dp,-1,sizeof(dp));dp[0]=0;
	printf("%d",DP(S)+ans);
	return 0;
}
### Prompt

Please create a solution in CPP 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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int c[101000],ok[10000000];
int h[101000],nxt[201000],K=0,to[201000],cost[201000];int cnt[16],ans=0;
void ins(int u,int v,int c){nxt[++K]=h[u];h[u]=K;to[K]=v;cost[K]=c;}
void dfs(int x,int f,int val)
{
	int k=0;
	for(int i=h[x];i;i=nxt[i])
	{
		if(to[i]==f)continue;
		k^=cost[i];dfs(to[i],x,cost[i]);
	}
	if(x)cnt[val^k]++;
}
int a[18],tot=0,dp[1<<15];
int DP(int S)
{
	
	if(dp[S]!=-1)return dp[S];dp[S]=2000000000;
	for(int i=0;i<15;i++)
	for(int j=i+1;j<15;j++)
	{
		if(((S>>i)&1)&&((S>>j)&1))
		{
			int k=(i+1)^(j+1);
			dp[S]=min(dp[S],DP(S^(1<<(k-1))^(1<<j)^(1<<i))+1+(((S>>(k-1))&1)?1:0));
		}
	}
	for(int i=0;i<15;i++)
		if((S>>i)&1)
			dp[S]=min(dp[S],DP(S^(1<<i))+1);
	return dp[S];
}
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<n;i++){int x,y,c;scanf("%d%d%d",&x,&y,&c);ins(x,y,c);ins(y,x,c);}
	dfs(0,-1,0);
	int S=0;
	for(int i=1;i<=15;i++)
	{
		ans+=cnt[i]/2;
		if(cnt[i]&1)S|=1<<(i-1);
	}
	memset(dp,-1,sizeof(dp));dp[0]=0;
	printf("%d",DP(S)+ans);
	return 0;
}
```