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
#include<bits/stdc++.h>
#define testbit(mask,i) (((mask)>>(i))&1)
#define setbit(mask,i) ((mask)|(1<<i))
#define delbit(mask,i) ((mask)^(1<<i))
#define randome_generate srand(std::chrono::high_resolution_clock::now().time_since_epoch().count())
using namespace std;
vector<pair<int,int> > ga[100100];
int u,v,w;
int n;
int cnt[18];
int dq;
int xo[262144],dp[262144];
int main(){
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n;
	for(int i=1;i<n;i++){
		cin>>u>>v>>w;
		ga[u].push_back(make_pair(v,w));
		ga[v].push_back(make_pair(u,w));
	}
	for(int i=0;i<n;i++){
		dq=0;
		for(int j=0;j<ga[i].size();j++){
			dq^=ga[i][j].second;
		} 
		cnt[dq]++;
	}
	int cc=0;
	cc+=cnt[0];
	int zt=0;
	for(int i=1;i<=15;i++){
		cc+=cnt[i]/2;
		cnt[i]%=2;
		if(cnt[i])
			zt+=(1<<i);
	}
	for(int i=0;i<(1<<16);i++){
		for(int j=0;j<=15;j++){
			if(testbit(i,j)){
				xo[i]^=j;
			}
		}
	}
	for(int i=0;i<(1<<16);i++){
		for(int j=i;j>0;j=(j-1)&i){
			if(xo[j]==0)
				dp[i]=max(dp[i],dp[i^j]+1);
		}
	}
	cc+=dp[zt];
	cout<<n-cc<<endl;
	return 0;
}
