#include<bits/stdc++.h>
#define ll long long
#define mid (l+r)/2
using namespace std;
const int INF=0x3f3f3f3f;
const int MOD=1e5+3;
const int N=2e5+7;
int read()
{
    int f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}
void print(int x)
{
    if(x<0){putchar('-');x=-x;}
    if(x>9)print(x/10);
    putchar(x%10+'0');
}
int n,m,k;
int a[N],f[N],cnt[N],xo[N],ans=0,sit,up;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;++i)
	{
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		a[u]^=w;a[v]^=w;
	}
	up=(1<<15);
	for(int i=0;i<n;++i)
		cnt[a[i]]++;
	for(int i=1;i<=15;++i)
	{
		ans+=cnt[i]/2;
		sit|=((cnt[i]&1)<<(i-1));
	}
	for(int i=1;i<up;++i)
		for(int j=0;j<15;++j)
			if((i>>j)&1) xo[i]^=(j+1); 
	for(int i=1;i<up;++i)
		f[i]=f[i>>1]+(i&1);//统计1的个数（f的上界）
	for(int i=1;i<up;++i)
		f[i]--;
	for(int i=1;i<up;++i)
	{
		if(xo[i]) continue;
		for(int k=(i-1)&i;k;k=(k-1)&i)
		{
			if(xo[k]) continue;
			f[i]=min(f[i],f[k]+f[i^k]);
		}
	}
	printf("%d",ans+f[sit]);
	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>
#define ll long long
#define mid (l+r)/2
using namespace std;
const int INF=0x3f3f3f3f;
const int MOD=1e5+3;
const int N=2e5+7;
int read()
{
    int f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}
void print(int x)
{
    if(x<0){putchar('-');x=-x;}
    if(x>9)print(x/10);
    putchar(x%10+'0');
}
int n,m,k;
int a[N],f[N],cnt[N],xo[N],ans=0,sit,up;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;++i)
	{
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		a[u]^=w;a[v]^=w;
	}
	up=(1<<15);
	for(int i=0;i<n;++i)
		cnt[a[i]]++;
	for(int i=1;i<=15;++i)
	{
		ans+=cnt[i]/2;
		sit|=((cnt[i]&1)<<(i-1));
	}
	for(int i=1;i<up;++i)
		for(int j=0;j<15;++j)
			if((i>>j)&1) xo[i]^=(j+1); 
	for(int i=1;i<up;++i)
		f[i]=f[i>>1]+(i&1);//统计1的个数（f的上界）
	for(int i=1;i<up;++i)
		f[i]--;
	for(int i=1;i<up;++i)
	{
		if(xo[i]) continue;
		for(int k=(i-1)&i;k;k=(k-1)&i)
		{
			if(xo[k]) continue;
			f[i]=min(f[i],f[k]+f[i^k]);
		}
	}
	printf("%d",ans+f[sit]);
	return 0;
}
```