#pragma GCC optimize("Ofast,unroll-loops")
#include<bits/stdc++.h>
#define ll long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++) 
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
int c[16],n,v[200005],dp[1<<16];
int dfs(int x){
	if(dp[x]!=-1)RE dp[x];
	int re=1000000000;
	rep(i,1,16){
		if(x&(1<<i)){
			rep(j,i+1,16){
				if(x&(1<<j)){
					re=min(re,dfs(x^(1<<i)^(1<<j)^(1<<(i^j)))+((x&(1<<(i^j)))?2:1));
				}
			}
			break;
		}
	}
	RE dp[x]=re;
}
int main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	cin>>n;
	int x,y,z;
	rep(i,1,n){
		cin>>x>>y>>z;
		v[x]^=z;v[y]^=z;
	}
	int ans=0;
	rep(i,0,n){
		c[v[i]]++;
	}
	FOR(i,1,15){
		ans+=c[i]/2;
		c[i]&=1;
	}
	int get=0;
	FILL(dp,-1);dp[0]=0;
	FOR(i,1,15)if(c[i])get^=(1<<i);
	cout<<dfs(get)+ans;
	RE 0;
}



### Prompt

Construct a cpp code solution to the problem outlined:

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
#pragma GCC optimize("Ofast,unroll-loops")
#include<bits/stdc++.h>
#define ll long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++) 
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
int c[16],n,v[200005],dp[1<<16];
int dfs(int x){
	if(dp[x]!=-1)RE dp[x];
	int re=1000000000;
	rep(i,1,16){
		if(x&(1<<i)){
			rep(j,i+1,16){
				if(x&(1<<j)){
					re=min(re,dfs(x^(1<<i)^(1<<j)^(1<<(i^j)))+((x&(1<<(i^j)))?2:1));
				}
			}
			break;
		}
	}
	RE dp[x]=re;
}
int main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	cin>>n;
	int x,y,z;
	rep(i,1,n){
		cin>>x>>y>>z;
		v[x]^=z;v[y]^=z;
	}
	int ans=0;
	rep(i,0,n){
		c[v[i]]++;
	}
	FOR(i,1,15){
		ans+=c[i]/2;
		c[i]&=1;
	}
	int get=0;
	FILL(dp,-1);dp[0]=0;
	FOR(i,1,15)if(c[i])get^=(1<<i);
	cout<<dfs(get)+ans;
	RE 0;
}



```