#include<bits/stdc++.h>
using namespace std;
int n,x,y,z,ans,f[1<<15],a[100100],c[20];
vector<int>o;

int main(){
	ios::sync_with_stdio(false);
	scanf("%d", &n);
	for(int i=1;i<n;++i){
		scanf("%d%d%d",&x,&y,&z);
		a[x]^=z,a[y]^=z;
	}
	for(int i=0;i<n;++i)++c[a[i]];
	for(int i=1;i<=15;++i){
		ans+=c[i]>>1;
		if(c[i]&1)o.push_back(i);
	}
	f[0]=0;
	for(int i=1;i<(1<<o.size());++i){
		int x=0,k=0,j=0;
		for(;j<o.size();++j)
			if(i&(1<<j))x^=o[j],++k;
		f[i]=x? 19260817:k-1;
		for(j=i-(i&-i);j;j=(i-(i&-i))&(j-1))
			f[i]=min(f[i],f[j]+f[j^i]);
	}
	printf("%d\n",f[(1<<o.size())-1]+ans);
	return 0;
}
### Prompt

Your task is to create 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>
using namespace std;
int n,x,y,z,ans,f[1<<15],a[100100],c[20];
vector<int>o;

int main(){
	ios::sync_with_stdio(false);
	scanf("%d", &n);
	for(int i=1;i<n;++i){
		scanf("%d%d%d",&x,&y,&z);
		a[x]^=z,a[y]^=z;
	}
	for(int i=0;i<n;++i)++c[a[i]];
	for(int i=1;i<=15;++i){
		ans+=c[i]>>1;
		if(c[i]&1)o.push_back(i);
	}
	f[0]=0;
	for(int i=1;i<(1<<o.size());++i){
		int x=0,k=0,j=0;
		for(;j<o.size();++j)
			if(i&(1<<j))x^=o[j],++k;
		f[i]=x? 19260817:k-1;
		for(j=i-(i&-i);j;j=(i-(i&-i))&(j-1))
			f[i]=min(f[i],f[j]+f[j^i]);
	}
	printf("%d\n",f[(1<<o.size())-1]+ans);
	return 0;
}
```