#include <bits/stdc++.h>
#define MAXN 100005
#define MAXM 65537
#define INF 0x3f3f3f3f
//#define int long long
using namespace std;
inline int read(){
	int x=0,f=1;
	char ch=getchar();
	while (ch<'0'||ch>'9'){
		if (ch=='-') f=-1;
		ch=getchar();
	}
	while (ch>='0'&&ch<='9'){
		x=(x*10)+(ch-'0');
		ch=getchar();
	}
	return x*f;
}
int dp[MAXM];
int val[MAXN],cnt[MAXN];
int dfs(int s){
	//printf("s:%d\n",s);
	if (s==0) return dp[s]=0;
	if (dp[s]!=INF) return dp[s];
	for (register int i=0;i<16;++i){
		if (!(s&(1<<i))) continue;
		for (register int j=0;j<16;++j){
			if (i==j||(!(s&(1<<j)))) continue;
			int p=i^j;
			dp[s]=min(dp[s],dfs(s^(1<<i)^(1<<j)^(1<<p))+1+((s&(1<<p))?1:0));
		}
	}
	return dp[s];
}
//#undef int
int main(){
//#define int long long
	int n=read();
	for (register int i=1;i<n;++i){
		int u=read()+1,v=read()+1,w=read();
		val[u]^=w,val[v]^=w;
	}
	for (register int i=1;i<=n;++i){
		cnt[val[i]]++;
	}
	int ans=0,s=0;
	for (register int i=1;i<16;++i){
		ans+=cnt[i]>>1;
		if (cnt[i]&1) s|=(1<<i);
	}
	//printf("OK\n");
	memset(dp,0x3f,sizeof(dp));
	printf("%d\n",ans+dfs(s));
}
### Prompt

Please create a solution in cpp 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>
#define MAXN 100005
#define MAXM 65537
#define INF 0x3f3f3f3f
//#define int long long
using namespace std;
inline int read(){
	int x=0,f=1;
	char ch=getchar();
	while (ch<'0'||ch>'9'){
		if (ch=='-') f=-1;
		ch=getchar();
	}
	while (ch>='0'&&ch<='9'){
		x=(x*10)+(ch-'0');
		ch=getchar();
	}
	return x*f;
}
int dp[MAXM];
int val[MAXN],cnt[MAXN];
int dfs(int s){
	//printf("s:%d\n",s);
	if (s==0) return dp[s]=0;
	if (dp[s]!=INF) return dp[s];
	for (register int i=0;i<16;++i){
		if (!(s&(1<<i))) continue;
		for (register int j=0;j<16;++j){
			if (i==j||(!(s&(1<<j)))) continue;
			int p=i^j;
			dp[s]=min(dp[s],dfs(s^(1<<i)^(1<<j)^(1<<p))+1+((s&(1<<p))?1:0));
		}
	}
	return dp[s];
}
//#undef int
int main(){
//#define int long long
	int n=read();
	for (register int i=1;i<n;++i){
		int u=read()+1,v=read()+1,w=read();
		val[u]^=w,val[v]^=w;
	}
	for (register int i=1;i<=n;++i){
		cnt[val[i]]++;
	}
	int ans=0,s=0;
	for (register int i=1;i<16;++i){
		ans+=cnt[i]>>1;
		if (cnt[i]&1) s|=(1<<i);
	}
	//printf("OK\n");
	memset(dp,0x3f,sizeof(dp));
	printf("%d\n",ans+dfs(s));
}
```