#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define fi first
#define se second
#define debug(x) cerr << #x << " " << x << '\n'
using namespace std;
using ll = long long;
using pii = pair<int,int>;
using pli = pair<ll,int>;
const int INF = 0x3f3f3f3f, N = 1e5 + 5;
const ll LINF = 1e18 + 5;
constexpr int mod = 1e9 + 7;
int n, dp[1<<16], a[N], cnt[16];
int dfs(int S)
{
	if(!S) return 0;
	if(~dp[S]) return dp[S];
	int &x = dp[S]; x = INF;
	for(int i=0; i<=15; i++) if((S>>i)&1)
		for(int j=0; j<=15; j++) if(i^j && (S>>j)&1)
		{
			int k = i^j;
			int T = S^(1<<i)^(1<<j)^(1<<k);
			if((T>>k)&1) x = min(x, dfs(T)+1);
			else x = min(x, dfs(T)+2);
		}
	return x;
}
int main()
{
	scanf("%d", &n);
	for(int i=1; i<n; i++)
	{
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z);
		a[x] ^= z; a[y] ^= z;
	}
	for(int i=0; i<n; i++) cnt[a[i]]++;
	memset(dp, -1, sizeof(dp));
	int S = 0, ans = 0;
	for(int i=1; i<=15; i++)
	{
		ans += cnt[i]/2;
		if(cnt[i]&1) S |= (1<<i);
	}
	printf("%d\n", dfs(S) + ans);
	return 0;
}
### Prompt

Your task is to create 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>
#define mp make_pair
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define fi first
#define se second
#define debug(x) cerr << #x << " " << x << '\n'
using namespace std;
using ll = long long;
using pii = pair<int,int>;
using pli = pair<ll,int>;
const int INF = 0x3f3f3f3f, N = 1e5 + 5;
const ll LINF = 1e18 + 5;
constexpr int mod = 1e9 + 7;
int n, dp[1<<16], a[N], cnt[16];
int dfs(int S)
{
	if(!S) return 0;
	if(~dp[S]) return dp[S];
	int &x = dp[S]; x = INF;
	for(int i=0; i<=15; i++) if((S>>i)&1)
		for(int j=0; j<=15; j++) if(i^j && (S>>j)&1)
		{
			int k = i^j;
			int T = S^(1<<i)^(1<<j)^(1<<k);
			if((T>>k)&1) x = min(x, dfs(T)+1);
			else x = min(x, dfs(T)+2);
		}
	return x;
}
int main()
{
	scanf("%d", &n);
	for(int i=1; i<n; i++)
	{
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z);
		a[x] ^= z; a[y] ^= z;
	}
	for(int i=0; i<n; i++) cnt[a[i]]++;
	memset(dp, -1, sizeof(dp));
	int S = 0, ans = 0;
	for(int i=1; i<=15; i++)
	{
		ans += cnt[i]/2;
		if(cnt[i]&1) S |= (1<<i);
	}
	printf("%d\n", dfs(S) + ans);
	return 0;
}
```