#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;

const int N=100005;

int ans,n,x,y,w,m,sz;
int dp[N],b[N],a[N],cnt[N];
bool tag[N];

void init(){
	int s,i,ss;
	sz=(1<<m)-1;
	for (s=0;s<=sz;s++){
		ss=0;
		for (i=1;i<=m;i++) if ((s>>i-1)&1) ss^=a[i];
		if (!ss) tag[s]=1;
	}
}

int main(){
	int i,s,ss,is;
	scanf("%d",&n);
	for (i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&w); x++; y++;
		b[x]^=w; b[y]^=w;
	}
	for (i=1;i<=n;i++) cnt[b[i]]++;
	ans+=cnt[0];
	for (i=1;i<=15;i++){
		ans+=cnt[i]/2;
		if (cnt[i]&1) a[++m]=i;
	}
	init();
	for (s=0;s<=sz;s++){
		is=sz^s;
		for (ss=is;ss;ss=(ss-1)&is)
			if (tag[ss]) dp[s|ss]=max(dp[s|ss],dp[s]+1);
	}
	printf("%d\n",n-(ans+dp[sz]));
	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>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;

const int N=100005;

int ans,n,x,y,w,m,sz;
int dp[N],b[N],a[N],cnt[N];
bool tag[N];

void init(){
	int s,i,ss;
	sz=(1<<m)-1;
	for (s=0;s<=sz;s++){
		ss=0;
		for (i=1;i<=m;i++) if ((s>>i-1)&1) ss^=a[i];
		if (!ss) tag[s]=1;
	}
}

int main(){
	int i,s,ss,is;
	scanf("%d",&n);
	for (i=1;i<n;i++){
		scanf("%d%d%d",&x,&y,&w); x++; y++;
		b[x]^=w; b[y]^=w;
	}
	for (i=1;i<=n;i++) cnt[b[i]]++;
	ans+=cnt[0];
	for (i=1;i<=15;i++){
		ans+=cnt[i]/2;
		if (cnt[i]&1) a[++m]=i;
	}
	init();
	for (s=0;s<=sz;s++){
		is=sz^s;
		for (ss=is;ss;ss=(ss-1)&is)
			if (tag[ss]) dp[s|ss]=max(dp[s|ss],dp[s]+1);
	}
	printf("%d\n",n-(ans+dp[sz]));
	return 0;
}
```