#include <stdio.h>

int N,X[100100],C[1<<16],D[1<<16];

int main()
{
	scanf ("%d",&N);
	for (int i=1;i<N;i++){
		int x,y,c; scanf ("%d %d %d",&x,&y,&c);
		X[x] ^= c;
		X[y] ^= c;
	}

	for (int i=0;i<N;i++) C[1<<X[i]]++;
	C[1] = 0;

	int ans = 16;
	for (int i=1;i<16;i++){
		int &c = C[1<<i];
		ans += c / 2;
		c = c % 2 * i;
	}

	for (int i=1;i<(1<<16);i++){
		int x = i & (-i);
		C[i] = C[i-x] ^ C[x];

		D[i] = -99999;
		for (int j=i;j;j=(j-1)&i) if (C[j] == 0){
			if (D[i] < D[i-j] + 1)
				D[i] = D[i-j] + 1;
		}
	}
	ans -= D[(1<<16)-1];

	printf ("%d\n",ans);

	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 <stdio.h>

int N,X[100100],C[1<<16],D[1<<16];

int main()
{
	scanf ("%d",&N);
	for (int i=1;i<N;i++){
		int x,y,c; scanf ("%d %d %d",&x,&y,&c);
		X[x] ^= c;
		X[y] ^= c;
	}

	for (int i=0;i<N;i++) C[1<<X[i]]++;
	C[1] = 0;

	int ans = 16;
	for (int i=1;i<16;i++){
		int &c = C[1<<i];
		ans += c / 2;
		c = c % 2 * i;
	}

	for (int i=1;i<(1<<16);i++){
		int x = i & (-i);
		C[i] = C[i-x] ^ C[x];

		D[i] = -99999;
		for (int j=i;j;j=(j-1)&i) if (C[j] == 0){
			if (D[i] < D[i-j] + 1)
				D[i] = D[i-j] + 1;
		}
	}
	ans -= D[(1<<16)-1];

	printf ("%d\n",ans);

	return 0;
}
```