#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MN 200005
using namespace std;
int n,m,a[MN],val[MN],ans,f[1<<16],sum[20];
int main(){
	scanf("%d",&n);int x,y,v;
	for(int i=1;i<n;i++)scanf("%d%d%d",&x,&y,&v),a[x]^=v,a[y]^=v;
	for(int i=0;i<n;i++)sum[a[i]]++;
	for(int i=1;i<=15;i++){
		ans+=sum[i]/2;
		if(sum[i]%2)val[m++]=i;
	}memset(f,0x3f,sizeof(f));f[0]=0;
	for(int i=1;i<(1<<m);i++){
		int x=0,cnt=0;
		for(int j=0;j<m;j++)if(i&(1<<j))x^=val[j],cnt++;
		if(!x)f[i]=cnt-1;
		for(int y=i-(i&(-i)),z=y;z;z=(z-1)&y,f[i]=min(f[i],f[z]+f[i^z]));
	}printf("%d\n",f[(1<<m)-1]+ans);
}

### Prompt

In CPP, your task is to solve 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<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MN 200005
using namespace std;
int n,m,a[MN],val[MN],ans,f[1<<16],sum[20];
int main(){
	scanf("%d",&n);int x,y,v;
	for(int i=1;i<n;i++)scanf("%d%d%d",&x,&y,&v),a[x]^=v,a[y]^=v;
	for(int i=0;i<n;i++)sum[a[i]]++;
	for(int i=1;i<=15;i++){
		ans+=sum[i]/2;
		if(sum[i]%2)val[m++]=i;
	}memset(f,0x3f,sizeof(f));f[0]=0;
	for(int i=1;i<(1<<m);i++){
		int x=0,cnt=0;
		for(int j=0;j<m;j++)if(i&(1<<j))x^=val[j],cnt++;
		if(!x)f[i]=cnt-1;
		for(int y=i-(i&(-i)),z=y;z;z=(z-1)&y,f[i]=min(f[i],f[z]+f[i^z]));
	}printf("%d\n",f[(1<<m)-1]+ans);
}

```