#include <bits/stdc++.h>
#define MAXN 100007
using namespace std;
vector<int> g[MAXN],w[MAXN],v;
int sga[MAXN],dp[MAXN],cnt[16];
void dfs(int s,int f,int d)
{
	int x=0;
	for(int i=0;i<g[s].size();i++) if(g[s][i]!=f) dfs(g[s][i],s,w[s][i]);
	for(int i=0;i<g[s].size();i++) if(g[s][i]!=f) x^=w[s][i];
	if(x!=d && s!=f) cnt[x^d]++;
}
int main()
{
	int n; cin>>n;
	for(int i=0;i<n-1;i++)
	{
		int t1,t2,t3;
		cin>>t1>>t2>>t3;
		g[t1].push_back(t2);
		g[t2].push_back(t1);
		w[t1].push_back(t3);
		w[t2].push_back(t3);
	}
	dfs(0,0,0);
	int sol=0;
	for(int i=1;i<=15;i++) {sol+=cnt[i]/2; if(cnt[i]&1) v.push_back(i);}
	int m=v.size();
	for(int i=0;i<m;i++) sga[1<<i]=1;
	for(int i=0;i<(1<<m);i++)
	{
		int x=0;
		for(int j=0;j<m;j++) if((1<<j)&i) x^=v[j];
		if(x==0) sga[i]=__builtin_popcount(i)-1;
	}
	fill(dp,dp+MAXN,100000); dp[0]=0;
	for(int i=1;i<(1<<m);i++)
	{
		int sb=i;
		do
		{
			if(sga[sb]) dp[i]=min(dp[i],dp[i^sb]+sga[sb]);
        	sb = (sb-1) & i;
    	} while(sb!=0);
	}
	cout<<sol+dp[(1<<m)-1];
}

### Prompt

In cpp, your task is to solve 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 100007
using namespace std;
vector<int> g[MAXN],w[MAXN],v;
int sga[MAXN],dp[MAXN],cnt[16];
void dfs(int s,int f,int d)
{
	int x=0;
	for(int i=0;i<g[s].size();i++) if(g[s][i]!=f) dfs(g[s][i],s,w[s][i]);
	for(int i=0;i<g[s].size();i++) if(g[s][i]!=f) x^=w[s][i];
	if(x!=d && s!=f) cnt[x^d]++;
}
int main()
{
	int n; cin>>n;
	for(int i=0;i<n-1;i++)
	{
		int t1,t2,t3;
		cin>>t1>>t2>>t3;
		g[t1].push_back(t2);
		g[t2].push_back(t1);
		w[t1].push_back(t3);
		w[t2].push_back(t3);
	}
	dfs(0,0,0);
	int sol=0;
	for(int i=1;i<=15;i++) {sol+=cnt[i]/2; if(cnt[i]&1) v.push_back(i);}
	int m=v.size();
	for(int i=0;i<m;i++) sga[1<<i]=1;
	for(int i=0;i<(1<<m);i++)
	{
		int x=0;
		for(int j=0;j<m;j++) if((1<<j)&i) x^=v[j];
		if(x==0) sga[i]=__builtin_popcount(i)-1;
	}
	fill(dp,dp+MAXN,100000); dp[0]=0;
	for(int i=1;i<(1<<m);i++)
	{
		int sb=i;
		do
		{
			if(sga[sb]) dp[i]=min(dp[i],dp[i^sb]+sga[sb]);
        	sb = (sb-1) & i;
    	} while(sb!=0);
	}
	cout<<sol+dp[(1<<m)-1];
}

```