#include<bits/stdc++.h>
#define N 16
using namespace std;
int n,x,y,z,ans,val[1<<(N+1)],cnt[N*2],dp[1<<N];
vector<int>G[N*2];
int main(){
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&z);
		x++;y++;
		val[x]^=z;val[y]^=z;
	}
	for (int i=1;i<=n;i++) cnt[val[i]]++;
	for (int i=1;i<=15;i++) ans+=cnt[i]/2,val[i]=cnt[i]%2;
	int now=0,tot=0;
	for (int i=1;i<=15;i++) if (val[i]) now|=(1<<i),tot++;
	memset(dp,127,sizeof(dp));int inf=dp[now];dp[now]=0;
	for (int i=0;i<(1<<16);i++) G[__builtin_popcount(i)].push_back(i);
	for (int i=tot;i;i--){
		for (int j=0;j<(int)G[i].size();j++){
			int S=G[i][j];
			if (dp[S]==inf) continue;
			for (int x=1;x<=15;x++){
				for (int y=1;y<=15;y++){
					if (x==y||(!(S>>x&1))||(!(S>>y&1))) continue;
					int tmp=x^y;
					int S1=S^(1<<x)^(1<<y);
					if (S1&(1<<tmp)) dp[S1^(1<<tmp)]=min(dp[S1^(1<<tmp)],dp[S]+2);
					else dp[S1|(1<<tmp)]=min(dp[S1|(1<<tmp)],dp[S]+1);
				}
			}
		}
	}
	//cerr<<inf<<endl;
	printf("%d\n",min(dp[1],dp[0])+ans);
	return 0;
}
### Prompt

Generate 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 N 16
using namespace std;
int n,x,y,z,ans,val[1<<(N+1)],cnt[N*2],dp[1<<N];
vector<int>G[N*2];
int main(){
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&z);
		x++;y++;
		val[x]^=z;val[y]^=z;
	}
	for (int i=1;i<=n;i++) cnt[val[i]]++;
	for (int i=1;i<=15;i++) ans+=cnt[i]/2,val[i]=cnt[i]%2;
	int now=0,tot=0;
	for (int i=1;i<=15;i++) if (val[i]) now|=(1<<i),tot++;
	memset(dp,127,sizeof(dp));int inf=dp[now];dp[now]=0;
	for (int i=0;i<(1<<16);i++) G[__builtin_popcount(i)].push_back(i);
	for (int i=tot;i;i--){
		for (int j=0;j<(int)G[i].size();j++){
			int S=G[i][j];
			if (dp[S]==inf) continue;
			for (int x=1;x<=15;x++){
				for (int y=1;y<=15;y++){
					if (x==y||(!(S>>x&1))||(!(S>>y&1))) continue;
					int tmp=x^y;
					int S1=S^(1<<x)^(1<<y);
					if (S1&(1<<tmp)) dp[S1^(1<<tmp)]=min(dp[S1^(1<<tmp)],dp[S]+2);
					else dp[S1|(1<<tmp)]=min(dp[S1|(1<<tmp)],dp[S]+1);
				}
			}
		}
	}
	//cerr<<inf<<endl;
	printf("%d\n",min(dp[1],dp[0])+ans);
	return 0;
}
```