#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
int n,ans,k,t,b[100005],c[16],d[1<<16];
void gao(int i){
	for(int j=0;j<16;j++)
		if(i>>j&1)k^=j;
	if(!k){
		d[i]=1;
		for(int j=(i-1)&i;j;j=(j-1)&i)
			d[i]=max(d[i],d[j]+d[j^i]);
	}
}
void gg(int i){
	for(int j=0;j<16;j++)
		for(int p=0;p<16;p++)
			if(j&p>=i)k^=j&p;
}
int main(){
	n=read();
	for(int i=0;i<n-1;i++){
		int x=read(),y=read(),a=read();
		b[x]^=a,b[y]^=a;
	}
	for(int i=0;i<n;i++)
		c[b[i]]++;
	for(int i=1;i<1<<16;i++){
		k=0;
		gao(i);
	}
	for(int i=1;i<16;i++){
		ans+=c[i]/2;
		if(c[i]&1)t^=1<<i;
	}
	printf("%d",ans+__builtin_popcount(t)-d[t]);
	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<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
int n,ans,k,t,b[100005],c[16],d[1<<16];
void gao(int i){
	for(int j=0;j<16;j++)
		if(i>>j&1)k^=j;
	if(!k){
		d[i]=1;
		for(int j=(i-1)&i;j;j=(j-1)&i)
			d[i]=max(d[i],d[j]+d[j^i]);
	}
}
void gg(int i){
	for(int j=0;j<16;j++)
		for(int p=0;p<16;p++)
			if(j&p>=i)k^=j&p;
}
int main(){
	n=read();
	for(int i=0;i<n-1;i++){
		int x=read(),y=read(),a=read();
		b[x]^=a,b[y]^=a;
	}
	for(int i=0;i<n;i++)
		c[b[i]]++;
	for(int i=1;i<1<<16;i++){
		k=0;
		gao(i);
	}
	for(int i=1;i<16;i++){
		ans+=c[i]/2;
		if(c[i]&1)t^=1<<i;
	}
	printf("%d",ans+__builtin_popcount(t)-d[t]);
	return 0;
}
```