#include<bits/stdc++.h>
using namespace std;
const int inf=1e9,mod=1e9+7;
typedef long long ll;
int read() {
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9') f=(c=='-')?-1:1,c=getchar();
	while(c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}
using namespace std;
int val[1000001],f[1000001],cnt[20],vis[1000001],js,x,y,z,ans,n,len=(1<<15);
int main() {
	n=read(),len=(1<<15);
	for(int i=1; i<n; i++)
		x=read(),y=read(),z=read(),val[x]^=z,val[y]^=z;
	for(int i=0; i<n; i++)
		cnt[val[i]]++;
	for(int i=1; i<=15; i++)
		ans+=cnt[i]/2,js|=(cnt[i]&1)<<(i-1);
	for(int i=1; i<len; i++)
		f[i]=f[i>>1]+(i&1);
	for(int i=1; i<len; i++)
		f[i]--;
	for(int i=1; i<len; i++)
		for(int j=0; j<15; j++)
			if((i>>j)&1)
				vis[i]^=(j+1);
	for(int i=1; i<len; i++) {
		if(vis[i]) continue;
		for(int k=(i-1)&i; k; k=(k-1)&i)
			if(!vis[k])
				f[i]=min(f[i],f[k]+f[i^k]);
	}
	printf("%d",ans+f[js]);
	return 0;
}//da
### 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>
using namespace std;
const int inf=1e9,mod=1e9+7;
typedef long long ll;
int read() {
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9') f=(c=='-')?-1:1,c=getchar();
	while(c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}
using namespace std;
int val[1000001],f[1000001],cnt[20],vis[1000001],js,x,y,z,ans,n,len=(1<<15);
int main() {
	n=read(),len=(1<<15);
	for(int i=1; i<n; i++)
		x=read(),y=read(),z=read(),val[x]^=z,val[y]^=z;
	for(int i=0; i<n; i++)
		cnt[val[i]]++;
	for(int i=1; i<=15; i++)
		ans+=cnt[i]/2,js|=(cnt[i]&1)<<(i-1);
	for(int i=1; i<len; i++)
		f[i]=f[i>>1]+(i&1);
	for(int i=1; i<len; i++)
		f[i]--;
	for(int i=1; i<len; i++)
		for(int j=0; j<15; j++)
			if((i>>j)&1)
				vis[i]^=(j+1);
	for(int i=1; i<len; i++) {
		if(vis[i]) continue;
		for(int k=(i-1)&i; k; k=(k-1)&i)
			if(!vis[k])
				f[i]=min(f[i],f[k]+f[i^k]);
	}
	printf("%d",ans+f[js]);
	return 0;
}//da
```