#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <iostream>
using namespace std;
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define pb push_back
#define fi first
#define se second
#define ll long long
#define sz(x) (int)(x).size()
const int mxn=1e5+10;
int b[mxn];
int cnt[16];
int a[20];
int sz=0;
int dp[mxn];
bool flag[mxn];
inline void check(int mask){
	int x=0;
	for(int i=0;i<sz;i++){
		if(mask &(1<<i))x^=a[i];
	}
	flag[mask]= (x==0);
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		int u,v,w;
		scanf("%d %d %d",&u,&v,&w);
		b[u]^=w;
		b[v]^=w;
	}
	for(int i=0;i<n;i++){
		cnt[b[i]]++;
	}
	int component=cnt[0];
	for(int i=1;i<16;i++){
		component+=cnt[i]/2;
		if(cnt[i]%2)a[sz++]=i;
	}
	int up=(1<<sz)-1;
	for(int i=1;i<=up;i++){
		check(i);
		for(int x=i;x;x=((x-1)&i)){
			if(flag[i])dp[i]=max(dp[i],1+dp[x^i]);
		}
	}
	cout<<n-component-dp[up];
	return 0;
}
### Prompt

Please provide a cpp coded solution 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 <algorithm>
#include <cstring>
#include <vector>
#include <iostream>
using namespace std;
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define pb push_back
#define fi first
#define se second
#define ll long long
#define sz(x) (int)(x).size()
const int mxn=1e5+10;
int b[mxn];
int cnt[16];
int a[20];
int sz=0;
int dp[mxn];
bool flag[mxn];
inline void check(int mask){
	int x=0;
	for(int i=0;i<sz;i++){
		if(mask &(1<<i))x^=a[i];
	}
	flag[mask]= (x==0);
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		int u,v,w;
		scanf("%d %d %d",&u,&v,&w);
		b[u]^=w;
		b[v]^=w;
	}
	for(int i=0;i<n;i++){
		cnt[b[i]]++;
	}
	int component=cnt[0];
	for(int i=1;i<16;i++){
		component+=cnt[i]/2;
		if(cnt[i]%2)a[sz++]=i;
	}
	int up=(1<<sz)-1;
	for(int i=1;i<=up;i++){
		check(i);
		for(int x=i;x;x=((x-1)&i)){
			if(flag[i])dp[i]=max(dp[i],1+dp[x^i]);
		}
	}
	cout<<n-component-dp[up];
	return 0;
}
```