#include <cstdio>
#include <cctype>
#include <cstring>
const int maxn=100005;
int N,A[maxn],G[16],opt,F[65536],Ans;
inline void read(int &Res){
	char ch=getchar();
	for (Res=0;!isdigit(ch);ch=getchar());
	for (;isdigit(ch);ch=getchar()) Res=(Res<<3)+(Res<<1)+ch-48;
}
void DFS(int o,int w){
	if (w>=F[o]) return;
	F[o]=w;
	for (int i=1;i<16;++i) if ((o>>i)&1)
		for (int j=1;j<16;++j) if ((i^j)&&((o>>j)&1)) DFS(o^(1<<i)^(1<<j)^1^(1<<(i^j)),w+1+((o>>(i^j))&1));
}
int main(){
	read(N),memset(F,63,sizeof F);
	for (int i=1,x,y,z;i<N;++i) read(x),read(y),read(z),A[x]^=z,A[y]^=z;
	for (int i=0;i<N;++i) ++G[A[i]];
	for (int i=1;i<16;++i) Ans+=G[i]>>1,opt|=(G[i]&1)<<i;
	return DFS(opt,0),printf("%d",Ans+(*F<F[1]?*F:F[1])),0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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>
#include <cctype>
#include <cstring>
const int maxn=100005;
int N,A[maxn],G[16],opt,F[65536],Ans;
inline void read(int &Res){
	char ch=getchar();
	for (Res=0;!isdigit(ch);ch=getchar());
	for (;isdigit(ch);ch=getchar()) Res=(Res<<3)+(Res<<1)+ch-48;
}
void DFS(int o,int w){
	if (w>=F[o]) return;
	F[o]=w;
	for (int i=1;i<16;++i) if ((o>>i)&1)
		for (int j=1;j<16;++j) if ((i^j)&&((o>>j)&1)) DFS(o^(1<<i)^(1<<j)^1^(1<<(i^j)),w+1+((o>>(i^j))&1));
}
int main(){
	read(N),memset(F,63,sizeof F);
	for (int i=1,x,y,z;i<N;++i) read(x),read(y),read(z),A[x]^=z,A[y]^=z;
	for (int i=0;i<N;++i) ++G[A[i]];
	for (int i=1;i<16;++i) Ans+=G[i]>>1,opt|=(G[i]&1)<<i;
	return DFS(opt,0),printf("%d",Ans+(*F<F[1]?*F:F[1])),0;
}
```