#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100100;
int  n , deg[N] , cnt[16];
int ans , dp[1<<16] , msk;
int ok[1<<16];

int main(){

scanf("%d",&n);
for(int i=1;i<n;i++){
    int u,v,c;
    scanf("%d %d %d",&u,&v,&c);
    deg[u] ^= c;
    deg[v] ^= c;
}

for(int i=0;i<n;i++)cnt[deg[i]]++;
for(int i=1;i<16;i++){
    ans += (cnt[i]>>1);
    if( cnt[i]&1 )
        msk |= (1<<i);
}

for(int i=0;i<(1<<16);i++){
    int x =0;
    for(int j=0;j<16;j++)
        if( i&(1<<j) )
         x ^= j;
    ok[i]=(x==0);
}

for(int i=0;i<(1<<16);i++){
    int j = i ;
    while( j ){
        if( ok[j] )
            dp[i] = max( dp[i] , 1+dp[i^j] );
        j=(j-1)&i;
    }
}

printf("%d\n",ans+__builtin_popcount(msk)-dp[msk]);
return 0;
}

### Prompt

In cpp, your task is to solve 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100100;
int  n , deg[N] , cnt[16];
int ans , dp[1<<16] , msk;
int ok[1<<16];

int main(){

scanf("%d",&n);
for(int i=1;i<n;i++){
    int u,v,c;
    scanf("%d %d %d",&u,&v,&c);
    deg[u] ^= c;
    deg[v] ^= c;
}

for(int i=0;i<n;i++)cnt[deg[i]]++;
for(int i=1;i<16;i++){
    ans += (cnt[i]>>1);
    if( cnt[i]&1 )
        msk |= (1<<i);
}

for(int i=0;i<(1<<16);i++){
    int x =0;
    for(int j=0;j<16;j++)
        if( i&(1<<j) )
         x ^= j;
    ok[i]=(x==0);
}

for(int i=0;i<(1<<16);i++){
    int j = i ;
    while( j ){
        if( ok[j] )
            dp[i] = max( dp[i] , 1+dp[i^j] );
        j=(j-1)&i;
    }
}

printf("%d\n",ans+__builtin_popcount(msk)-dp[msk]);
return 0;
}

```