#include<bits/stdc++.h>
#define N 200005
using namespace std;
int bitcount(int S) {
	int ret = 0;
	while(S) ret ++, S -= (S & (-S));
	return ret;
}
int check(int S) {
	int ret = 0;
	for(int i = 0; i < 15; i ++) if(S & (1 << i)) ret ^= (i + 1);
	return ret;
}
int n, d[N], cnt[N], f[N];
int main() {
	scanf("%d", &n);
	for(int i = 1; i < n; i ++) {
		int u, v, c;
		scanf("%d%d%d", &u, &v, &c);
		d[u] ^= c, d[v] ^= c;
	}
	for(int i = 0; i < n; i ++) cnt[d[i]] ++;
	int ans = 0, sta = 0;
	for(int i = 1; i < 16; i ++) ans += cnt[i] / 2, sta |= (cnt[i]&1) << (i - 1);
	memset(f, 0x3f, sizeof f);
	f[0] = 0;
	for(int S = 1; S < (1 << 15); S ++) { 
		if((sta & S) != S || check(S)) continue;
		f[S] = bitcount(S) - 1;
		for(int SS = S; SS; SS = (SS - 1) & S) if(!check(SS)){ 
			f[S] = min(f[S], f[SS] + f[S ^ SS]);
		}
	}
	int ret = 0;
	printf("%d", ans + f[sta]);
	return 0;
} 
### Prompt

Create a solution in CPP for 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>
#define N 200005
using namespace std;
int bitcount(int S) {
	int ret = 0;
	while(S) ret ++, S -= (S & (-S));
	return ret;
}
int check(int S) {
	int ret = 0;
	for(int i = 0; i < 15; i ++) if(S & (1 << i)) ret ^= (i + 1);
	return ret;
}
int n, d[N], cnt[N], f[N];
int main() {
	scanf("%d", &n);
	for(int i = 1; i < n; i ++) {
		int u, v, c;
		scanf("%d%d%d", &u, &v, &c);
		d[u] ^= c, d[v] ^= c;
	}
	for(int i = 0; i < n; i ++) cnt[d[i]] ++;
	int ans = 0, sta = 0;
	for(int i = 1; i < 16; i ++) ans += cnt[i] / 2, sta |= (cnt[i]&1) << (i - 1);
	memset(f, 0x3f, sizeof f);
	f[0] = 0;
	for(int S = 1; S < (1 << 15); S ++) { 
		if((sta & S) != S || check(S)) continue;
		f[S] = bitcount(S) - 1;
		for(int SS = S; SS; SS = (SS - 1) & S) if(!check(SS)){ 
			f[S] = min(f[S], f[SS] + f[S ^ SS]);
		}
	}
	int ret = 0;
	printf("%d", ans + f[sta]);
	return 0;
} 
```