Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

5
0 1 1
0 2 3
0 3 6
3 4 4

Sample Output 1

3

The objective can be achieved in three operations, as follows:

First, choose the path connecting Vertex 1, 2, and x = 1.

Then, choose the path connecting Vertex 2, 3, and x = 2.

Lastly, choose the path connecting Vertex 0, 4, and x = 4.

Sample Input 2

2
1 0 0

Sample Output 2

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> pii;
const int maxn=100005;
int n,x[maxn],a[16],dp[1<<16],s,mask[1<<16];
vector<int> v;
vector<pii> G[maxn];
void dfs(int u,int fa) {
    for (int i=0;i<(int)G[u].size();++i) {
        int v=G[u][i].first,w=G[u][i].second;
        if (v==fa)
            continue;
        x[v]=w^x[u];
        dfs(v,u);
    }
}
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n-1;++i) {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        x[u]^=w;
        x[v]^=w;
        G[u].push_back(pii(v,w));
        G[v].push_back(pii(u,w));
    }
    //dfs(0,-1);
    for (int i=0;i<n;++i)
        ++a[x[i]];
    int res=n-a[0];
    for (int i=1;i<16;++i) {
        if (a[i]%2==1)
            v.push_back(i);
        res-=a[i]/2;
    }
    if (!v.size())
        return 0*printf("%d\n",res);
    s=1<<v.size();
    for (int i=0;i<s;++i)
        for (int j=0;j<(int)v.size();++j)
            if (i&(1<<j))
                mask[i]^=v[j];
    for (int i=0;i<s;++i)
        for (int j=0;j<(int)v.size();++j)
            if (i&(1<<j))
                dp[i]=max(dp[i],dp[i^(1<<j)]+(mask[i]==0));
    printf("%d\n",res-dp[s-1]);
    return 0;
}