#include <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef pair<lint, int> pi;
typedef long long lint;
const int MAXN = 100005;

int n;
int cnt[16], arr[MAXN];
int dp[MAXN];

int bfs(int msk){
	for(int i=1; i<(1<<16); i++){
		int ans = 0;
		dp[i] = -1e9;
		for(int j=0; j<16; j++){
			if((i >> j) & 1) ans ^= j;
		}
		if(ans == 0) dp[i] = 1;
		for(int j=i; j>0; j=(j-1)&i){
			dp[i] = max(dp[j] + dp[i^j], dp[i]);
		}
	}
	int ans = 0;
	for(int i=0; i<16; i++) if((msk >> i) & 1) ans++;
	return ans - dp[msk];
}

int main(){
	scanf("%d",&n);
	for(int i=0; i<n-1; i++){
		int s, e, x;
		scanf("%d %d %d",&s,&e,&x);
		arr[s] ^= x;
		arr[e] ^= x;
	}
	for(int i=0; i<n; i++) cnt[arr[i]]++;// printf("%d\n", arr[i]);
	int msk = 0, ans = 0;
	for(int i=1; i<16; i++){
		ans += cnt[i] / 2;
		if(cnt[i] & 1) msk |= (1 << i);
	}
	cout << ans + bfs(msk) << endl;
}

### Prompt

Please create a solution in Cpp 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 <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef pair<lint, int> pi;
typedef long long lint;
const int MAXN = 100005;

int n;
int cnt[16], arr[MAXN];
int dp[MAXN];

int bfs(int msk){
	for(int i=1; i<(1<<16); i++){
		int ans = 0;
		dp[i] = -1e9;
		for(int j=0; j<16; j++){
			if((i >> j) & 1) ans ^= j;
		}
		if(ans == 0) dp[i] = 1;
		for(int j=i; j>0; j=(j-1)&i){
			dp[i] = max(dp[j] + dp[i^j], dp[i]);
		}
	}
	int ans = 0;
	for(int i=0; i<16; i++) if((msk >> i) & 1) ans++;
	return ans - dp[msk];
}

int main(){
	scanf("%d",&n);
	for(int i=0; i<n-1; i++){
		int s, e, x;
		scanf("%d %d %d",&s,&e,&x);
		arr[s] ^= x;
		arr[e] ^= x;
	}
	for(int i=0; i<n; i++) cnt[arr[i]]++;// printf("%d\n", arr[i]);
	int msk = 0, ans = 0;
	for(int i=1; i<16; i++){
		ans += cnt[i] / 2;
		if(cnt[i] & 1) msk |= (1 << i);
	}
	cout << ans + bfs(msk) << endl;
}

```