#include<cstdio>
#include<vector>
using namespace std;
const int maxn=100000,inf=0x3f3f3f3f;
struct edge{int to,len;};
vector<edge> g[maxn+10];
int n,a[maxn+10],ans,c[16],ccnt,f[1<<16],gg[1<<16],h[1<<16]; bool b[16];
int main(){
	scanf("%d",&n);
	for(int i=1;i<n;++i){
		int l,r,w; scanf("%d%d%d",&l,&r,&w);
		a[l+1]^=w; a[r+1]^=w;
	}
	for(int i=1;i<=n;++i) if(a[i])
		if(b[a[i]]) ++f[0],b[a[i]]=0; else b[a[i]]=1;
	for(int i=0;i<16;++i) if(b[i]) c[ccnt++]=i;
	for(int i=0;i<1<<ccnt;++i)
		for(int j=0;j<ccnt;++j) if(i>>j&1) gg[i]^=c[j],++h[i];
	for(int i=1;i<1<<ccnt;++i){
		f[i]=inf;
		for(int j=i;j;j=(j-1)&i) if(!gg[j]) f[i]=min(f[i],f[i^j]+h[j]-1);
	}
	printf("%d",f[(1<<ccnt)-1]);
}
### 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<cstdio>
#include<vector>
using namespace std;
const int maxn=100000,inf=0x3f3f3f3f;
struct edge{int to,len;};
vector<edge> g[maxn+10];
int n,a[maxn+10],ans,c[16],ccnt,f[1<<16],gg[1<<16],h[1<<16]; bool b[16];
int main(){
	scanf("%d",&n);
	for(int i=1;i<n;++i){
		int l,r,w; scanf("%d%d%d",&l,&r,&w);
		a[l+1]^=w; a[r+1]^=w;
	}
	for(int i=1;i<=n;++i) if(a[i])
		if(b[a[i]]) ++f[0],b[a[i]]=0; else b[a[i]]=1;
	for(int i=0;i<16;++i) if(b[i]) c[ccnt++]=i;
	for(int i=0;i<1<<ccnt;++i)
		for(int j=0;j<ccnt;++j) if(i>>j&1) gg[i]^=c[j],++h[i];
	for(int i=1;i<1<<ccnt;++i){
		f[i]=inf;
		for(int j=i;j;j=(j-1)&i) if(!gg[j]) f[i]=min(f[i],f[i^j]+h[j]-1);
	}
	printf("%d",f[(1<<ccnt)-1]);
}
```