#include <bits/stdc++.h>
#ifdef DEMETRIO
#define deb(...) fprintf(stderr,__VA_ARGS__)
#define deb1(x) cerr << #x << " = " << x << endl
#else
#define deb(...) 0
#define deb1(x) 0
#endif
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
#define SZ(x) ((int)x.size())
using namespace std;
typedef long long ll;

int n,r;
int x[100005];
int dp[1<<16];

int f(int m){
	int& r=dp[m];
	if(r>=0)return r;
	r=1<<30;
	fore(i,1,16)if(m&(1<<i))fore(j,1,16)if(j!=i&&(m&(1<<j))){
		int k=i^j;
		int nm=m^(1<<i)^(1<<j);
		int a=1;
		if(k){
			if(nm&(1<<k))a++;
			nm^=1<<k;
		}
		r=min(r,a+f(nm));
	}
	return r;
}

int main(){
	scanf("%d",&n);
	fore(_,1,n){
		int i,j,a;
		scanf("%d%d%d",&i,&j,&a);
		x[i]^=a;x[j]^=a;
	}
	int m=0;
	fore(i,0,n){
		if(!x[i])continue;
		if(m&(1<<x[i]))r++;
		m^=1<<x[i];
	}
	memset(dp,-1,sizeof(dp));dp[0]=0;
	printf("%d\n",r+f(m));
	return 0;
}
### Prompt

Please formulate 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>
#ifdef DEMETRIO
#define deb(...) fprintf(stderr,__VA_ARGS__)
#define deb1(x) cerr << #x << " = " << x << endl
#else
#define deb(...) 0
#define deb1(x) 0
#endif
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
#define SZ(x) ((int)x.size())
using namespace std;
typedef long long ll;

int n,r;
int x[100005];
int dp[1<<16];

int f(int m){
	int& r=dp[m];
	if(r>=0)return r;
	r=1<<30;
	fore(i,1,16)if(m&(1<<i))fore(j,1,16)if(j!=i&&(m&(1<<j))){
		int k=i^j;
		int nm=m^(1<<i)^(1<<j);
		int a=1;
		if(k){
			if(nm&(1<<k))a++;
			nm^=1<<k;
		}
		r=min(r,a+f(nm));
	}
	return r;
}

int main(){
	scanf("%d",&n);
	fore(_,1,n){
		int i,j,a;
		scanf("%d%d%d",&i,&j,&a);
		x[i]^=a;x[j]^=a;
	}
	int m=0;
	fore(i,0,n){
		if(!x[i])continue;
		if(m&(1<<x[i]))r++;
		m^=1<<x[i];
	}
	memset(dp,-1,sizeof(dp));dp[0]=0;
	printf("%d\n",r+f(m));
	return 0;
}
```