#include <bits/stdc++.h>
using namespace std;
#define mod 998244353
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fll
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int maxn=1e5+10;
const int maxt=(1<<16)+10;
int val[maxn];
int cnt[20];
vector<int> g[maxn];
int n,ans,hmsk;
int dp[maxt];
bool ok[maxt];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int a,b,e;
		scanf("%d%d%d",&a,&b,&e);
		g[a].push_back(b);
		g[b].push_back(a);
		val[a]^=e;
		val[b]^=e;
	}
	for(int i=0;i<n;i++)
	{
		if(!val[i])continue;
		cnt[val[i]]^=1;
		if(cnt[val[i]]==0)ans++;
	}
	for(int i=0;i<16;i++)
		if(cnt[i])hmsk|=(1<<i),ans++;
	for(int msk=0;msk<(1<<16);msk++)
	{
		int val=0;
		for(int i=0;i<16;i++)
		{
			if(msk>>i&1)val^=i;
		}
		if(val==0)ok[msk]=1;
	}
	for(int msk=0;msk<(1<<16);msk+=2)
	{
		for(int s=msk;s;s=(s-1)&msk)
		{
			if(!ok[s])continue;
			dp[msk]=max(dp[msk],dp[msk^s]+1);
		}
	}
	printf("%d\n",ans-dp[hmsk]);
	return 0;
}
### Prompt

Please create a solution in Cpp 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;
#define mod 998244353
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fll
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int maxn=1e5+10;
const int maxt=(1<<16)+10;
int val[maxn];
int cnt[20];
vector<int> g[maxn];
int n,ans,hmsk;
int dp[maxt];
bool ok[maxt];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int a,b,e;
		scanf("%d%d%d",&a,&b,&e);
		g[a].push_back(b);
		g[b].push_back(a);
		val[a]^=e;
		val[b]^=e;
	}
	for(int i=0;i<n;i++)
	{
		if(!val[i])continue;
		cnt[val[i]]^=1;
		if(cnt[val[i]]==0)ans++;
	}
	for(int i=0;i<16;i++)
		if(cnt[i])hmsk|=(1<<i),ans++;
	for(int msk=0;msk<(1<<16);msk++)
	{
		int val=0;
		for(int i=0;i<16;i++)
		{
			if(msk>>i&1)val^=i;
		}
		if(val==0)ok[msk]=1;
	}
	for(int msk=0;msk<(1<<16);msk+=2)
	{
		for(int s=msk;s;s=(s-1)&msk)
		{
			if(!ok[s])continue;
			dp[msk]=max(dp[msk],dp[msk^s]+1);
		}
	}
	printf("%d\n",ans-dp[hmsk]);
	return 0;
}
```