#include<iostream>
#include<cstdio>
const int N=1e5+5;
int n,dp[1<<16],tot[20]={},a[N]={};bool vis[1<<16]={};
int len(int x) {
    int res=0;for(int i=1;i<16;i++) if(x>>i&1) res++;return res;
}
int dfs(int x) {
    if(x<=1) return 0;
    if(len(x)==1) return N;
    if(vis[x]) return dp[x];
    int res=N;vis[x]=true;
    for(int i=1;i<16;i++) {
        if(~x>>i&1) continue;
        for(int j=1;j<16;j++) {
            if(i==j||(~x>>j&1)) continue;
            int nxt=x^(1<<i)^(1<<j)^(1<<(i^j));
            if(len(nxt)>=len(x)) continue;
            res=std::min(res,dfs(nxt)+1+(~nxt>>(i^j)&1));
        }
    }
    return dp[x]=res;
}
int main() {
    scanf("%d",&n);int x,y,z;
    for(int i=1;i<n;i++) scanf("%d%d%d",&x,&y,&z),a[x]^=z,a[y]^=z;
    for(int i=0;i<n;i++) tot[a[i]]++;
    int ans=0,sta=0;
    for(int i=1;i<16;i++) ans+=(tot[i]>>1),sta|=(tot[i]&1)<<i;
    printf("%d\n",ans+dfs(sta));
    return 0;
}

### Prompt

Your challenge is to write 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<iostream>
#include<cstdio>
const int N=1e5+5;
int n,dp[1<<16],tot[20]={},a[N]={};bool vis[1<<16]={};
int len(int x) {
    int res=0;for(int i=1;i<16;i++) if(x>>i&1) res++;return res;
}
int dfs(int x) {
    if(x<=1) return 0;
    if(len(x)==1) return N;
    if(vis[x]) return dp[x];
    int res=N;vis[x]=true;
    for(int i=1;i<16;i++) {
        if(~x>>i&1) continue;
        for(int j=1;j<16;j++) {
            if(i==j||(~x>>j&1)) continue;
            int nxt=x^(1<<i)^(1<<j)^(1<<(i^j));
            if(len(nxt)>=len(x)) continue;
            res=std::min(res,dfs(nxt)+1+(~nxt>>(i^j)&1));
        }
    }
    return dp[x]=res;
}
int main() {
    scanf("%d",&n);int x,y,z;
    for(int i=1;i<n;i++) scanf("%d%d%d",&x,&y,&z),a[x]^=z,a[y]^=z;
    for(int i=0;i<n;i++) tot[a[i]]++;
    int ans=0,sta=0;
    for(int i=1;i<16;i++) ans+=(tot[i]>>1),sta|=(tot[i]&1)<<i;
    printf("%d\n",ans+dfs(sta));
    return 0;
}

```