#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#define ll long long
#define N 100010
#define TOP (1<<15)
using namespace std;

ll n,t,ans,a[N],sum[N],f[N],Xor[N];

inline ll read(){
	ll x=0,tmp=1;
	char ch=getchar();
	while(!isdigit(ch)){
		if(ch=='-') tmp=-1;
		ch=getchar();
	}
	while(isdigit(ch)){
		x=(x<<3)+(x<<1)+(ch^48);
		ch=getchar();
	}
	return tmp*x;
}

int main(){
	n=read();
	for(ll i=1; i<n; i++){
		ll x=read()+1,y=read()+1,z=read();
		a[x]^=z; a[y]^=z;
	}
	for(ll i=1; i<=n; i++) sum[a[i]]++;
	for(ll i=1; i<=15; i++){
		ans+=sum[i]>>1;
		t|=(sum[i]&1)<<(i-1);
	}
    for(ll i=1; i<TOP; i++) f[i]=f[i>>1]+(i&1);
    for(ll i=1; i<TOP; i++){
    	f[i]--;
    	for(ll j=0; j<15; j++){
        	if((i>>j)&1) Xor[i]^=j+1;
		}
	}
    for(ll i=1; i<TOP; i++){
        if(Xor[i]) continue;
        for(ll j=(i-1)&i; j; j=(j-1)&i){
        	if(!Xor[j]) f[i]=min(f[i],f[j]+f[i^j]);
		}   
    }
	cout<<ans+f[t]<<endl;
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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>
#include<iostream>
#include<cstdio>
#define ll long long
#define N 100010
#define TOP (1<<15)
using namespace std;

ll n,t,ans,a[N],sum[N],f[N],Xor[N];

inline ll read(){
	ll x=0,tmp=1;
	char ch=getchar();
	while(!isdigit(ch)){
		if(ch=='-') tmp=-1;
		ch=getchar();
	}
	while(isdigit(ch)){
		x=(x<<3)+(x<<1)+(ch^48);
		ch=getchar();
	}
	return tmp*x;
}

int main(){
	n=read();
	for(ll i=1; i<n; i++){
		ll x=read()+1,y=read()+1,z=read();
		a[x]^=z; a[y]^=z;
	}
	for(ll i=1; i<=n; i++) sum[a[i]]++;
	for(ll i=1; i<=15; i++){
		ans+=sum[i]>>1;
		t|=(sum[i]&1)<<(i-1);
	}
    for(ll i=1; i<TOP; i++) f[i]=f[i>>1]+(i&1);
    for(ll i=1; i<TOP; i++){
    	f[i]--;
    	for(ll j=0; j<15; j++){
        	if((i>>j)&1) Xor[i]^=j+1;
		}
	}
    for(ll i=1; i<TOP; i++){
        if(Xor[i]) continue;
        for(ll j=(i-1)&i; j; j=(j-1)&i){
        	if(!Xor[j]) f[i]=min(f[i],f[j]+f[i^j]);
		}   
    }
	cout<<ans+f[t]<<endl;
	return 0;
}
```