#include <bits/stdc++.h>
using namespace std;
const int maxn=100000+10;
int n,a[maxn],cnt[20],f[1<<15],g[1<<15];

int main()
{
	scanf("%d",&n);
	int x,y,w;
	for(int i=1;i<n;i++)
	{
		scanf("%d%d%d",&x,&y,&w);
		a[x]^=w;a[y]^=w;
	}
	for(int i=0;i<n;i++) cnt[a[i]]++;
	int res=0,s=0;
	for(int i=1;i<=15;i++)
		res+=cnt[i]>>1,s|=(cnt[i]&1)<<(i-1);
	for(int i=0;i<(1<<15);i++)
		for(int j=0;j<15;j++)
			if(i&(1<<j)) f[i]++,g[i]^=j+1;
	for(int i=1;i<(1<<15);i++) f[i]=!g[i]?f[i]-1:0;
	for(int i=1;i<(1<<15);i++)
		if(!g[i])
		{
			for(int k=(i-1)&i;k;k=(k-1)&i)
				if(!g[k]) f[i]=min(f[i],f[k]+f[i^k]);
		}
	printf("%d\n",res+f[s]);
	return 0;
}
### 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 <bits/stdc++.h>
using namespace std;
const int maxn=100000+10;
int n,a[maxn],cnt[20],f[1<<15],g[1<<15];

int main()
{
	scanf("%d",&n);
	int x,y,w;
	for(int i=1;i<n;i++)
	{
		scanf("%d%d%d",&x,&y,&w);
		a[x]^=w;a[y]^=w;
	}
	for(int i=0;i<n;i++) cnt[a[i]]++;
	int res=0,s=0;
	for(int i=1;i<=15;i++)
		res+=cnt[i]>>1,s|=(cnt[i]&1)<<(i-1);
	for(int i=0;i<(1<<15);i++)
		for(int j=0;j<15;j++)
			if(i&(1<<j)) f[i]++,g[i]^=j+1;
	for(int i=1;i<(1<<15);i++) f[i]=!g[i]?f[i]-1:0;
	for(int i=1;i<(1<<15);i++)
		if(!g[i])
		{
			for(int k=(i-1)&i;k;k=(k-1)&i)
				if(!g[k]) f[i]=min(f[i],f[k]+f[i^k]);
		}
	printf("%d\n",res+f[s]);
	return 0;
}
```