#include<bits/stdc++.h>
using namespace std;
#define N 1000005
int ans,a[N],x,y,v,n,cc[17],c[N],q,mx[N],dp[N],ff[N];
int i,j,k;
int main(){
	cin>>n;
	for(i=1;i<n;++i){
		cin>>x>>y>>v;
		a[x]^=v;
		a[y]^=v;
	}
	for(i=0;i<n;++i)++cc[a[i]];
	for(i=1;i<=15;++i){
		ans+=cc[i]/2;
		q+=((cc[i]&1)<<i-1);
	}
	for(i=1;i<(1<<15);++i){
		dp[i]=dp[i>>1]+(i&1);
	}
	for(i=1;i<(1<<15);++i)--dp[i];
	for(i=1;i<(1<<15);++i){
		for(j=0;j<15;++j){
			if(((1<<j)&i))ff[i]^=(j+1);
		}
	}
	for(i=1;i<=n;++i){
		if(ff[i])continue;
		for(k=(i-1);k;k=(k-1)&i){
			if(ff[k]==0)dp[i]=min(dp[i],dp[k]+dp[i^k]);
		}
	}cout<<dp[q]+ans;
}
### Prompt

In CPP, your task is to solve 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;
#define N 1000005
int ans,a[N],x,y,v,n,cc[17],c[N],q,mx[N],dp[N],ff[N];
int i,j,k;
int main(){
	cin>>n;
	for(i=1;i<n;++i){
		cin>>x>>y>>v;
		a[x]^=v;
		a[y]^=v;
	}
	for(i=0;i<n;++i)++cc[a[i]];
	for(i=1;i<=15;++i){
		ans+=cc[i]/2;
		q+=((cc[i]&1)<<i-1);
	}
	for(i=1;i<(1<<15);++i){
		dp[i]=dp[i>>1]+(i&1);
	}
	for(i=1;i<(1<<15);++i)--dp[i];
	for(i=1;i<(1<<15);++i){
		for(j=0;j<15;++j){
			if(((1<<j)&i))ff[i]^=(j+1);
		}
	}
	for(i=1;i<=n;++i){
		if(ff[i])continue;
		for(k=(i-1);k;k=(k-1)&i){
			if(ff[k]==0)dp[i]=min(dp[i],dp[k]+dp[i^k]);
		}
	}cout<<dp[q]+ans;
}
```