#include<bits/stdc++.h>
using namespace std;
int n,val[100009],ans,mask,cnt[16],good[1<<16],dp[1<<16];
int main() {
	scanf("%d",&n);
	for(int i=0;i<n-1;i++) {
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		val[a]^=c;
		val[b]^=c;
	}
	for(int i=0;i<n;i++) cnt[val[i]]++;
	for(int i=1;i<16;i++) {
		ans+=cnt[i]>>1;
		if(cnt[i]&1) mask|=1<<i;
	}
	for(int i=1;i<1<<16;i++) {
		int j=0;
		for(int k=0;k<16;k++)
			if(1<<k&i)
				j^=k;
		good[i]=j==0;
	}
	for(int i=2;i<1<<16;i+=2)
		for(int j=i;j;j=(j-1)&i)
			if(good[j])
				dp[i]=max(dp[i],dp[i^j]+1);
	printf("%d\n",ans+__builtin_popcount(mask)-dp[mask]);
	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;
int n,val[100009],ans,mask,cnt[16],good[1<<16],dp[1<<16];
int main() {
	scanf("%d",&n);
	for(int i=0;i<n-1;i++) {
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		val[a]^=c;
		val[b]^=c;
	}
	for(int i=0;i<n;i++) cnt[val[i]]++;
	for(int i=1;i<16;i++) {
		ans+=cnt[i]>>1;
		if(cnt[i]&1) mask|=1<<i;
	}
	for(int i=1;i<1<<16;i++) {
		int j=0;
		for(int k=0;k<16;k++)
			if(1<<k&i)
				j^=k;
		good[i]=j==0;
	}
	for(int i=2;i<1<<16;i+=2)
		for(int j=i;j;j=(j-1)&i)
			if(good[j])
				dp[i]=max(dp[i],dp[i^j]+1);
	printf("%d\n",ans+__builtin_popcount(mask)-dp[mask]);
	return 0;
}
```