#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
#include <iostream>
#include <algorithm>
#include <cstdio>
#define N 100020
#define M 16
using namespace std;

int a[N];
bool u[N];
int f[1 << M | 1], g[1 << M | 1];

int main(void)
{
    int n, u, v, w;
    int i, j, o;

    scanf("%d", &n);
    for(i = 1; i < n; i ++)
    {
        scanf("%d %d %d", &u, &v, &w);
        a[u] ^= w;
        a[v] ^= w;
    }
    for(i = o = 0; i < n; i ++)
        if(a[i])
        {
            o += ::u[a[i]];
            ::u[a[i]] ^= 1;
        }
    for(i = n = 0; i < M; i ++)
        if(::u[i])
            a[n ++] = i;

    // for(i=0;i<n;i++)cout<<a[i]<<' ';cout<<endl;

    for(i = 1; i < (1 << n); i ++)
    {
        f[i] = __builtin_popcount(i) - 1;
        g[i] = g[i & (i - 1)] ^ a[__builtin_ctz(i)];
        for(j = i & (i - 1); j; j = i & (j - 1))
            if(!g[j])
                f[i] = min(f[i], f[j] + f[i ^ j]);
    }
    printf("%d\n", o + f[(1 << n) - 1]);

    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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
#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
#include <iostream>
#include <algorithm>
#include <cstdio>
#define N 100020
#define M 16
using namespace std;

int a[N];
bool u[N];
int f[1 << M | 1], g[1 << M | 1];

int main(void)
{
    int n, u, v, w;
    int i, j, o;

    scanf("%d", &n);
    for(i = 1; i < n; i ++)
    {
        scanf("%d %d %d", &u, &v, &w);
        a[u] ^= w;
        a[v] ^= w;
    }
    for(i = o = 0; i < n; i ++)
        if(a[i])
        {
            o += ::u[a[i]];
            ::u[a[i]] ^= 1;
        }
    for(i = n = 0; i < M; i ++)
        if(::u[i])
            a[n ++] = i;

    // for(i=0;i<n;i++)cout<<a[i]<<' ';cout<<endl;

    for(i = 1; i < (1 << n); i ++)
    {
        f[i] = __builtin_popcount(i) - 1;
        g[i] = g[i & (i - 1)] ^ a[__builtin_ctz(i)];
        for(j = i & (i - 1); j; j = i & (j - 1))
            if(!g[j])
                f[i] = min(f[i], f[j] + f[i ^ j]);
    }
    printf("%d\n", o + f[(1 << n) - 1]);

    return 0;
}

```