#include<bits/stdc++.h>
using namespace std;
const int maxn=100100;
void chkmin(int &x,int y){
	x=x<y?x:y;
}
int beg[maxn],tto[maxn<<1],nex[maxn<<1],w[maxn<<1],e;
void putin(int s,int t,int v){
	tto[++e]=t;
	nex[e]=beg[s];
	beg[s]=e;
	w[e]=v;
}
int cnt[20],ans;
void dfs(int u,int fa,int fv){
	int sum=fv;
	for(int i=beg[u];i;i=nex[i]){
		if(tto[i]==fa) continue;
		dfs(tto[i],u,w[i]);
		sum^=w[i];
	}
	if(sum==0) return;
	ans+=cnt[sum];
	cnt[sum]^=1;
}
int dp[1<<15];
int popcnt[1<<15];
int main(){
//	freopen("F.in","r",stdin);
	int n;
	scanf("%d",&n);
	int s,t,v;
	for(int i=1;i<n;i++){
		scanf("%d%d%d",&s,&t,&v);
		s++,t++;
		putin(s,t,v);
		putin(t,s,v);
	}
	dfs(1,0,0);
	for(int i=1;i<(1<<15);i++)
		popcnt[i]=popcnt[i>>1]+(i&1);

	int val;
	for(int i=1;i<(1<<15);i++){
		dp[i]=popcnt[i];
		val=0;
		for(int j=0;j<15;j++)
			if(i&(1<<j))
				val^=j+1;
		if(val==0)
			dp[i]--;
	}
	for(int i=1;i<(1<<15);i++)
		for(int j=i;j;j=i&(j-1))
			chkmin(dp[i],dp[j]+dp[i^j]);
	val=0;
	for(int i=1;i<=15;i++)
		if(cnt[i])
			val|=1<<(i-1);
	ans=ans+dp[val];
	printf("%d\n",ans);
	return 0;
}

### 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>
using namespace std;
const int maxn=100100;
void chkmin(int &x,int y){
	x=x<y?x:y;
}
int beg[maxn],tto[maxn<<1],nex[maxn<<1],w[maxn<<1],e;
void putin(int s,int t,int v){
	tto[++e]=t;
	nex[e]=beg[s];
	beg[s]=e;
	w[e]=v;
}
int cnt[20],ans;
void dfs(int u,int fa,int fv){
	int sum=fv;
	for(int i=beg[u];i;i=nex[i]){
		if(tto[i]==fa) continue;
		dfs(tto[i],u,w[i]);
		sum^=w[i];
	}
	if(sum==0) return;
	ans+=cnt[sum];
	cnt[sum]^=1;
}
int dp[1<<15];
int popcnt[1<<15];
int main(){
//	freopen("F.in","r",stdin);
	int n;
	scanf("%d",&n);
	int s,t,v;
	for(int i=1;i<n;i++){
		scanf("%d%d%d",&s,&t,&v);
		s++,t++;
		putin(s,t,v);
		putin(t,s,v);
	}
	dfs(1,0,0);
	for(int i=1;i<(1<<15);i++)
		popcnt[i]=popcnt[i>>1]+(i&1);

	int val;
	for(int i=1;i<(1<<15);i++){
		dp[i]=popcnt[i];
		val=0;
		for(int j=0;j<15;j++)
			if(i&(1<<j))
				val^=j+1;
		if(val==0)
			dp[i]--;
	}
	for(int i=1;i<(1<<15);i++)
		for(int j=i;j;j=i&(j-1))
			chkmin(dp[i],dp[j]+dp[i^j]);
	val=0;
	for(int i=1;i<=15;i++)
		if(cnt[i])
			val|=1<<(i-1);
	ans=ans+dp[val];
	printf("%d\n",ans);
	return 0;
}

```