#include<bits/stdc++.h>
using namespace std;
inline int read(){
	int res=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-f;ch=getchar();}
	while(ch>='0'&&ch<='9'){res=(res<<3)+(res<<1)+ch-'0';ch=getchar();}
	return res*f;
}
void write(int x){
	if(x<0) putchar('-'),x=-x;
	if(x>9) write(x/10);
	putchar(x%10|'0');
}
const int N=100005;
int n,a[N],cnt[20],ans,st;
int f[1<<16],xs[1<<16];

int main(){
	n=read();
	for(int i=1;i<n;i++) {
		int u=read(),v=read(),w=read();
		a[u]^=w;a[v]^=w;
	}
	for(int i=0;i<n;i++) cnt[a[i]]++;
	for(int i=1;i<=15;i++){
		ans+=cnt[i]/2;st|=(cnt[i]&1)*(1<<(i-1));
	}
	int Lim=(1<<16)-1;
	for(int i=1;i<=Lim;i++) f[i]=f[i>>1]+(i&1); 
	for(int i=1;i<=Lim;i++) f[i]--; 
	for(int i=1;i<=Lim;i++) 
		for(int j=1;j<=15;j++)
			if(i&(1<<(j-1))){
				xs[i]=xs[i^(1<<(j-1))]^j;break;
			}
	for(int i=1;i<=st;i++){
		if(xs[i]) continue;
		for(int j=i&(i-1);j;j=(j-1)&i){
			if(xs[j]) continue;
			f[i]=min(f[i],f[j]+f[i^j]);
		}		
	}
	write(ans+f[st]);
	return 0;
}
### Prompt

Your challenge is to write 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;
inline int read(){
	int res=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-f;ch=getchar();}
	while(ch>='0'&&ch<='9'){res=(res<<3)+(res<<1)+ch-'0';ch=getchar();}
	return res*f;
}
void write(int x){
	if(x<0) putchar('-'),x=-x;
	if(x>9) write(x/10);
	putchar(x%10|'0');
}
const int N=100005;
int n,a[N],cnt[20],ans,st;
int f[1<<16],xs[1<<16];

int main(){
	n=read();
	for(int i=1;i<n;i++) {
		int u=read(),v=read(),w=read();
		a[u]^=w;a[v]^=w;
	}
	for(int i=0;i<n;i++) cnt[a[i]]++;
	for(int i=1;i<=15;i++){
		ans+=cnt[i]/2;st|=(cnt[i]&1)*(1<<(i-1));
	}
	int Lim=(1<<16)-1;
	for(int i=1;i<=Lim;i++) f[i]=f[i>>1]+(i&1); 
	for(int i=1;i<=Lim;i++) f[i]--; 
	for(int i=1;i<=Lim;i++) 
		for(int j=1;j<=15;j++)
			if(i&(1<<(j-1))){
				xs[i]=xs[i^(1<<(j-1))]^j;break;
			}
	for(int i=1;i<=st;i++){
		if(xs[i]) continue;
		for(int j=i&(i-1);j;j=(j-1)&i){
			if(xs[j]) continue;
			f[i]=min(f[i],f[j]+f[i^j]);
		}		
	}
	write(ans+f[st]);
	return 0;
}
```