#include<bits/stdc++.h>
using namespace std;
#define pb push_back
vector<pair<int,int> > g[400010];
int a[200010],w[200010],cnt[20],p[1000010],x,y,z,n,ans;
bool vis[1000010];
void dfs(int p,int fa)
{
	for(auto i:g[p])if(i.first!=fa)a[i.first]=i.second,dfs(i.first,p);
	for(auto i:g[p])if(i.first!=fa)w[p]^=w[i.first];
	a[p]^=w[p];w[p]^=a[p];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)scanf("%d%d%d",&x,&y,&z),x++,y++,g[x].pb({y,z}),g[y].pb({x,z});
	dfs(1,0);
	for(int i=1;i<=n;i++)++cnt[a[i]];
	for(int i=1;i<16;i++)ans+=cnt[i]/2,cnt[i]&=1;
	for(int i=1;i<16;i++)if(cnt[i])++ans;
	for(int i=1;i<(1<<16);i++)p[i]=-100;
	for(int i=0;i<(1<<15);i++)
	{
		int tp=i,ts=1,now=0;
		for(;tp;tp>>=1,++ts)if(tp&1)now^=ts;
		vis[i]=!now;
	}
	for(int i=1;i<(1<<15);i++)
	if(vis[i])
	{
		p[i]=1;
		for(int j=i;j;j=(j-1)&i)if(vis[j])p[i]=max(p[i],p[j]+p[i^j]);
	}
	int tp=0;
	for(int i=15;i;i--)tp=(tp<<1)|cnt[i];
	printf("%d\n",ans-p[tp]);
	return 0;
}
### Prompt

Create a solution in cpp for 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>
using namespace std;
#define pb push_back
vector<pair<int,int> > g[400010];
int a[200010],w[200010],cnt[20],p[1000010],x,y,z,n,ans;
bool vis[1000010];
void dfs(int p,int fa)
{
	for(auto i:g[p])if(i.first!=fa)a[i.first]=i.second,dfs(i.first,p);
	for(auto i:g[p])if(i.first!=fa)w[p]^=w[i.first];
	a[p]^=w[p];w[p]^=a[p];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)scanf("%d%d%d",&x,&y,&z),x++,y++,g[x].pb({y,z}),g[y].pb({x,z});
	dfs(1,0);
	for(int i=1;i<=n;i++)++cnt[a[i]];
	for(int i=1;i<16;i++)ans+=cnt[i]/2,cnt[i]&=1;
	for(int i=1;i<16;i++)if(cnt[i])++ans;
	for(int i=1;i<(1<<16);i++)p[i]=-100;
	for(int i=0;i<(1<<15);i++)
	{
		int tp=i,ts=1,now=0;
		for(;tp;tp>>=1,++ts)if(tp&1)now^=ts;
		vis[i]=!now;
	}
	for(int i=1;i<(1<<15);i++)
	if(vis[i])
	{
		p[i]=1;
		for(int j=i;j;j=(j-1)&i)if(vis[j])p[i]=max(p[i],p[j]+p[i^j]);
	}
	int tp=0;
	for(int i=15;i;i--)tp=(tp<<1)|cnt[i];
	printf("%d\n",ans-p[tp]);
	return 0;
}
```