#include<bits/stdc++.h>
using namespace std;
const int maxm=25;
const int maxs=(1<<15)+1;
const int maxn=2e5+5;
int n,ans,cnt;
int sum[maxs],pw[maxs],F[maxs],G[maxs],B[maxs];
int A[maxn],id[maxn];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n-1;++i){
		int a=0,b=0,c=0;
		scanf("%d%d%d",&a,&b,&c);
		A[a+1]^=c,A[b+1]^=c;
	}
	for(int i=1;i<=n;++i) ++sum[A[i]];
	for(int i=1;i<=15;++i){
		ans+=sum[i]/2;
		sum[i]%=2;
		if(sum[i]) B[++cnt]=i;
	}
	pw[0]=1;
	for(int i=1;i<=cnt;++i) pw[i]=pw[i-1]<<1,id[pw[i]]=i;
	for(int mask=1;mask<pw[cnt];++mask){
		F[mask]=10000;
		for(int i=mask;i;i-=i&-i){
			int res=mask^(i&-i),s=id[i&-i];
			G[mask]=G[res]^B[s+1];
			if(!G[res]) F[mask]=min(F[mask],F[res]);
			else F[mask]=min(F[mask],F[res]+1);
			//cout<<mask<<" "<<res<<" "<<B[s+1]<<" "<<s<<" "<<F[mask]<<" "<<G[mask]<<endl;
		}
	}
	printf("%d\n",ans+F[pw[cnt]-1]);
	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;
const int maxm=25;
const int maxs=(1<<15)+1;
const int maxn=2e5+5;
int n,ans,cnt;
int sum[maxs],pw[maxs],F[maxs],G[maxs],B[maxs];
int A[maxn],id[maxn];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n-1;++i){
		int a=0,b=0,c=0;
		scanf("%d%d%d",&a,&b,&c);
		A[a+1]^=c,A[b+1]^=c;
	}
	for(int i=1;i<=n;++i) ++sum[A[i]];
	for(int i=1;i<=15;++i){
		ans+=sum[i]/2;
		sum[i]%=2;
		if(sum[i]) B[++cnt]=i;
	}
	pw[0]=1;
	for(int i=1;i<=cnt;++i) pw[i]=pw[i-1]<<1,id[pw[i]]=i;
	for(int mask=1;mask<pw[cnt];++mask){
		F[mask]=10000;
		for(int i=mask;i;i-=i&-i){
			int res=mask^(i&-i),s=id[i&-i];
			G[mask]=G[res]^B[s+1];
			if(!G[res]) F[mask]=min(F[mask],F[res]);
			else F[mask]=min(F[mask],F[res]+1);
			//cout<<mask<<" "<<res<<" "<<B[s+1]<<" "<<s<<" "<<F[mask]<<" "<<G[mask]<<endl;
		}
	}
	printf("%d\n",ans+F[pw[cnt]-1]);
	return 0;
}
```