//It is made by M_sea
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define re register
using namespace std;

inline int read() {
    int X=0,w=1; char c=getchar();
    while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
    while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
    return X*w;
}

const int N=200000+10;

int w[N],cnt[20],f[1<<16],sxor[1<<16];

int main() {
    int n=read();
    for (re int i=1;i<n;++i) {
        int u=read()+1,v=read()+1,val=read();
        w[u]^=val,w[v]^=val;
    }
    for (re int i=1;i<=n;++i) ++cnt[w[i]];
    int ans=0,s=0;
    for (re int i=1;i<=15;++i) ans+=cnt[i]/2,s|=(cnt[i]&1)<<(i-1);
    for (re int i=1;i<(1<<15);++i) f[i]=f[i>>1]+(i&1);
    for (re int i=1;i<(1<<15);++i) --f[i];
    for (re int i=1;i<(1<<15);++i)
        for (re int j=1;j<=15;++j)
            if (i&(1<<(j-1))) sxor[i]^=j;
    for (re int i=1;i<(1<<15);++i) {
        if (sxor[i]) continue;
        for (re int j=i&(i-1);j;j=(j-1)&i)
            if (!sxor[j]) f[i]=min(f[i],f[j]+f[i^j]);
    }
    printf("%d\n",ans+f[s]);
    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
//It is made by M_sea
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define re register
using namespace std;

inline int read() {
    int X=0,w=1; char c=getchar();
    while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
    while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
    return X*w;
}

const int N=200000+10;

int w[N],cnt[20],f[1<<16],sxor[1<<16];

int main() {
    int n=read();
    for (re int i=1;i<n;++i) {
        int u=read()+1,v=read()+1,val=read();
        w[u]^=val,w[v]^=val;
    }
    for (re int i=1;i<=n;++i) ++cnt[w[i]];
    int ans=0,s=0;
    for (re int i=1;i<=15;++i) ans+=cnt[i]/2,s|=(cnt[i]&1)<<(i-1);
    for (re int i=1;i<(1<<15);++i) f[i]=f[i>>1]+(i&1);
    for (re int i=1;i<(1<<15);++i) --f[i];
    for (re int i=1;i<(1<<15);++i)
        for (re int j=1;j<=15;++j)
            if (i&(1<<(j-1))) sxor[i]^=j;
    for (re int i=1;i<(1<<15);++i) {
        if (sxor[i]) continue;
        for (re int j=i&(i-1);j;j=(j-1)&i)
            if (!sxor[j]) f[i]=min(f[i],f[j]+f[i^j]);
    }
    printf("%d\n",ans+f[s]);
    return 0;
}
```