#include<bits/stdc++.h>
using namespace std;

template<typename t>inline t min_(t a,t b){return a<=b?a:b;}
template<typename t>inline t max_(t a,t b){return a>=b?a:b;}
template<typename t>inline void chkmin(t&a,t b){if(a>b)a=b;}
template<typename t>inline void chkmax(t&a,t b){if(a<b)a=b;}

typedef long long ll;
const int maxn=1e5+10,maxa=16;
int n,a[maxn],cnt[maxa],ans;
int tot,x[maxa],dp[1<<maxa];
bool vis[1<<maxa];

int main(){
	cin>>n;
	for(int i=1,u,v,x;i<n;++i){
		cin>>u>>v>>x;
		a[u]^=x;a[v]^=x;
	}
	for(int i=0;i<n;++i)
		++cnt[a[i]];
	for(int i=1;i<maxa;++i)
		ans+=cnt[i]/2,cnt[i]-=cnt[i]/2*2;
	for(int i=1;i<maxa;++i)
		if(cnt[i])
			x[tot++]=i;
	for(int i=0;i<(1<<tot);++i){
		int xxx=0;
		for(int j=0;j<tot;++j)
			if(i>>j&1)
				xxx^=x[j];
		if(xxx==0)
			vis[i]=true;
	}
	memset(dp,100,sizeof(dp));
	dp[0]=0;
	for(int i=0;i<(1<<tot);++i)
		for(int j=(i+1)|i;j<(1<<tot);j=(j+1)|i)
			if(vis[i^j])
				chkmin(dp[j],dp[i]+__builtin_popcount(i^j)-1);
	printf("%d\n",ans+dp[(1<<tot)-1]);
	return 0;
}

### Prompt

Your challenge is to write 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<bits/stdc++.h>
using namespace std;

template<typename t>inline t min_(t a,t b){return a<=b?a:b;}
template<typename t>inline t max_(t a,t b){return a>=b?a:b;}
template<typename t>inline void chkmin(t&a,t b){if(a>b)a=b;}
template<typename t>inline void chkmax(t&a,t b){if(a<b)a=b;}

typedef long long ll;
const int maxn=1e5+10,maxa=16;
int n,a[maxn],cnt[maxa],ans;
int tot,x[maxa],dp[1<<maxa];
bool vis[1<<maxa];

int main(){
	cin>>n;
	for(int i=1,u,v,x;i<n;++i){
		cin>>u>>v>>x;
		a[u]^=x;a[v]^=x;
	}
	for(int i=0;i<n;++i)
		++cnt[a[i]];
	for(int i=1;i<maxa;++i)
		ans+=cnt[i]/2,cnt[i]-=cnt[i]/2*2;
	for(int i=1;i<maxa;++i)
		if(cnt[i])
			x[tot++]=i;
	for(int i=0;i<(1<<tot);++i){
		int xxx=0;
		for(int j=0;j<tot;++j)
			if(i>>j&1)
				xxx^=x[j];
		if(xxx==0)
			vis[i]=true;
	}
	memset(dp,100,sizeof(dp));
	dp[0]=0;
	for(int i=0;i<(1<<tot);++i)
		for(int j=(i+1)|i;j<(1<<tot);j=(j+1)|i)
			if(vis[i^j])
				chkmin(dp[j],dp[i]+__builtin_popcount(i^j)-1);
	printf("%d\n",ans+dp[(1<<tot)-1]);
	return 0;
}

```