#include<cstdio>
inline int Min(int x,int y){return x<y?x:y;}
int n,a[100000];
int s[16],Ans,S;
int f[1<<16];
int main(){
	int x,y,z;
	scanf("%d",&n);
	for(int i=1;i<n;++i) scanf("%d%d%d",&x,&y,&z), a[x]^=z, a[y]^=z;
	for(int i=0;i<n;++i) if(a[i]) Ans+=s[a[i]], s[a[i]]^=1;
	for(int i=1;i<16;++i) S^=s[i]<<i;
	for(int l=1;l<=15;++l) for(int i=1;i<1<<16;++i) if(__builtin_popcount(i)==l&&(i&1)==0){
		f[i]=(-1u)>>1;
		for(int j=1;j<16;++j) if(i>>j&1)
			for(int k=j+1;k<16;++k) if(i>>k&1)
				f[i]=Min(f[i],f[i^(1<<j)^(1<<k)^(1<<(j^k))]+1+(i>>(j^k)&1));
	}
	printf("%d",Ans+f[S]);
	return 0;
}
### Prompt

Your task is to create 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<cstdio>
inline int Min(int x,int y){return x<y?x:y;}
int n,a[100000];
int s[16],Ans,S;
int f[1<<16];
int main(){
	int x,y,z;
	scanf("%d",&n);
	for(int i=1;i<n;++i) scanf("%d%d%d",&x,&y,&z), a[x]^=z, a[y]^=z;
	for(int i=0;i<n;++i) if(a[i]) Ans+=s[a[i]], s[a[i]]^=1;
	for(int i=1;i<16;++i) S^=s[i]<<i;
	for(int l=1;l<=15;++l) for(int i=1;i<1<<16;++i) if(__builtin_popcount(i)==l&&(i&1)==0){
		f[i]=(-1u)>>1;
		for(int j=1;j<16;++j) if(i>>j&1)
			for(int k=j+1;k<16;++k) if(i>>k&1)
				f[i]=Min(f[i],f[i^(1<<j)^(1<<k)^(1<<(j^k))]+1+(i>>(j^k)&1));
	}
	printf("%d",Ans+f[S]);
	return 0;
}
```