#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read(){
	int Res=0,f=1;char ch=getchar();
	while (ch>'9'||ch<'0') f=(ch=='-'?-f:f),ch=getchar();
	while (ch>='0'&&ch<='9') Res=Res*10+ch-'0',ch=getchar();
	return Res*f;
}
int N,Ans,P,val[100005],F[32768];
bool vis[100005];
int dfs(int S){
	if (!S) return 0;
	if (F[S]) return F[S];
	int Res=1e9;
	for (int i=1;i<=15;i++){
		if (!((S>>i-1)&1)) continue;
		for (int j=1;j<=15;j++){
			if (!((S>>j-1)&1)) continue;
			if (i==j) continue;
			int Ns=S^(1<<i-1)^(1<<j-1)^(1<<(i^j)-1),add=1;
			if ((S>>(i^j)-1)&1) add++;
			Res=min(Res,dfs(Ns)+add);
		}
	}
	return F[S]=Res;
}
int main()
{
	N=read();
	for (int i=1;i<N;i++){
		int x=read(),y=read(),z=read();
		val[x]^=z;val[y]^=z;
	}
	for (int i=0;i<N;i++){if (vis[val[i]]&&val[i]) Ans++;vis[val[i]]^=1;}
	for (int i=1;i<=15;i++) P|=(vis[i]<<i-1);
	printf("%d\n",dfs(P)+Ans);
	return 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read(){
	int Res=0,f=1;char ch=getchar();
	while (ch>'9'||ch<'0') f=(ch=='-'?-f:f),ch=getchar();
	while (ch>='0'&&ch<='9') Res=Res*10+ch-'0',ch=getchar();
	return Res*f;
}
int N,Ans,P,val[100005],F[32768];
bool vis[100005];
int dfs(int S){
	if (!S) return 0;
	if (F[S]) return F[S];
	int Res=1e9;
	for (int i=1;i<=15;i++){
		if (!((S>>i-1)&1)) continue;
		for (int j=1;j<=15;j++){
			if (!((S>>j-1)&1)) continue;
			if (i==j) continue;
			int Ns=S^(1<<i-1)^(1<<j-1)^(1<<(i^j)-1),add=1;
			if ((S>>(i^j)-1)&1) add++;
			Res=min(Res,dfs(Ns)+add);
		}
	}
	return F[S]=Res;
}
int main()
{
	N=read();
	for (int i=1;i<N;i++){
		int x=read(),y=read(),z=read();
		val[x]^=z;val[y]^=z;
	}
	for (int i=0;i<N;i++){if (vis[val[i]]&&val[i]) Ans++;vis[val[i]]^=1;}
	for (int i=1;i<=15;i++) P|=(vis[i]<<i-1);
	printf("%d\n",dfs(P)+Ans);
	return 0;
}
```