#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const int MN = 100010;

int N;
int val[MN];
int num[16];
int dp[1 << 15];
int ret;
bool zero[1 << 15];

int main() {
	cin >> N;

	rep(i, N-1) {
		int a, b, e;
		cin >> a >> b >> e;
		val[a] ^= e;
		val[b] ^= e;
	}

	rep(i, N) num[val[i]]++;
	
	int f = 0;

	for (int i = 1; i < 16; ++i) {
		ret += num[i] / 2;
		num[i] %= 2;
		f |= (num[i] << (i-1));
	}

	rep(i, 1 << 15) {
		int xo = 0;
		rep(j, 15) {
			if ((i >> j) & 1) {
				xo ^= (j + 1);
			}
		}
		zero[i] = (xo == 0);
	}

	for (int i = 1; i < (1<<15); ++i) {
		for (int j = i; j >= 0; --j) {
			j &= i;
			if (j == 0) break;
			if (zero[j]) {
				dp[i] = max(dp[i], dp[i^j] + 1);
			}
		}
	}

	cout << ret + __builtin_popcount(f) - dp[f] << endl;

	return 0;

}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const int MN = 100010;

int N;
int val[MN];
int num[16];
int dp[1 << 15];
int ret;
bool zero[1 << 15];

int main() {
	cin >> N;

	rep(i, N-1) {
		int a, b, e;
		cin >> a >> b >> e;
		val[a] ^= e;
		val[b] ^= e;
	}

	rep(i, N) num[val[i]]++;
	
	int f = 0;

	for (int i = 1; i < 16; ++i) {
		ret += num[i] / 2;
		num[i] %= 2;
		f |= (num[i] << (i-1));
	}

	rep(i, 1 << 15) {
		int xo = 0;
		rep(j, 15) {
			if ((i >> j) & 1) {
				xo ^= (j + 1);
			}
		}
		zero[i] = (xo == 0);
	}

	for (int i = 1; i < (1<<15); ++i) {
		for (int j = i; j >= 0; --j) {
			j &= i;
			if (j == 0) break;
			if (zero[j]) {
				dp[i] = max(dp[i], dp[i^j] + 1);
			}
		}
	}

	cout << ret + __builtin_popcount(f) - dp[f] << endl;

	return 0;

}
```