#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define swap(a,b) (a^=b^=a^=b)
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
#define MN 100005
int n,val[MN],num[16],ans;
int f[1<<16],inf;
int dfs(int S)
{
	register int i,j;
	if(!S) return f[S]=0; 
	if(f[S]<inf)return f[S];
    for(i=0;i<16;i++)if(S&(1<<i))for(j=0;j<16;j++)if(i!=j&&(S&(1<<j)))
    {
    	int p=i^j,x=S^(1<<i)^(1<<j)^(1<<p);
    	if(S&(1<<p)) f[S]=min(f[S],dfs(x)+2);
    	else f[S]=min(f[S],dfs(x)+1);
    }
    return f[S];
}
int main()
{
	n=read();
	register int i,j,k,x,y,z,S=0;
	for(i=1;i<n;++i)
		x=read(),y=read(),z=read(),val[x]^=z,val[y]^=z;
	for(i=0;i<n;++i) num[val[i]]++;
	for(i=1;i<16;++i) ans+=(num[i]>>1),S+=(1<<i)*(num[i]&1);
	memset(f,127,sizeof(f));inf=f[0];f[0]=0;
	printf("%d\n",ans+dfs(S));
}
### Prompt

Your challenge is to write 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 max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define swap(a,b) (a^=b^=a^=b)
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
#define MN 100005
int n,val[MN],num[16],ans;
int f[1<<16],inf;
int dfs(int S)
{
	register int i,j;
	if(!S) return f[S]=0; 
	if(f[S]<inf)return f[S];
    for(i=0;i<16;i++)if(S&(1<<i))for(j=0;j<16;j++)if(i!=j&&(S&(1<<j)))
    {
    	int p=i^j,x=S^(1<<i)^(1<<j)^(1<<p);
    	if(S&(1<<p)) f[S]=min(f[S],dfs(x)+2);
    	else f[S]=min(f[S],dfs(x)+1);
    }
    return f[S];
}
int main()
{
	n=read();
	register int i,j,k,x,y,z,S=0;
	for(i=1;i<n;++i)
		x=read(),y=read(),z=read(),val[x]^=z,val[y]^=z;
	for(i=0;i<n;++i) num[val[i]]++;
	for(i=1;i<16;++i) ans+=(num[i]>>1),S+=(1<<i)*(num[i]&1);
	memset(f,127,sizeof(f));inf=f[0];f[0]=0;
	printf("%d\n",ans+dfs(S));
}
```