#include <cstdio>
#include <cstring>

const int MAXN = 1e5, INF = 0x3f3f3f3f;
int n, val[MAXN + 5], cnt[16], f[1 << 16], xsum[1 << 16];

inline void chkmin ( int& a, const int b ) { if ( b < a ) a = b; }

int main () {
	scanf ( "%d", &n );
	for ( int i = 1, u, v, w; i < n; ++ i ) {
		scanf ( "%d %d %d", &u, &v, &w );
		val[u] ^= w, val[v] ^= w;
	}
	int ans = 0, S = 0;
	for ( int i = 0; i < n; ++ i ) ++ cnt[val[i]];
	for ( int i = 1; i < 16; ++ i ) {
		S |= ( cnt[i] & 1 ) << i >> 1;
		ans += cnt[i] >> 1;
	}
	for ( int i = 1; i < 1 << 15; ++ i ) {
		for ( int j = 0; j < 15; ++ j ) {
			if ( ( i >> j ) & 1 ) {
				++ f[i], xsum[i] ^= j + 1;
			}
		}
		-- f[i];
	}
	for ( int s = 0; s < 1 << 15; ++ s ) {
		if ( xsum[s] ) continue;
		for ( int t = s; ; t = ( t - 1 ) & s ) {
			if ( ! xsum[t] && ! xsum[s ^ t] ) chkmin ( f[s], f[t] + f[s ^ t] );
			if ( ! t ) break;
		}
	}
	printf ( "%d\n", ans + f[S] );
	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 <cstdio>
#include <cstring>

const int MAXN = 1e5, INF = 0x3f3f3f3f;
int n, val[MAXN + 5], cnt[16], f[1 << 16], xsum[1 << 16];

inline void chkmin ( int& a, const int b ) { if ( b < a ) a = b; }

int main () {
	scanf ( "%d", &n );
	for ( int i = 1, u, v, w; i < n; ++ i ) {
		scanf ( "%d %d %d", &u, &v, &w );
		val[u] ^= w, val[v] ^= w;
	}
	int ans = 0, S = 0;
	for ( int i = 0; i < n; ++ i ) ++ cnt[val[i]];
	for ( int i = 1; i < 16; ++ i ) {
		S |= ( cnt[i] & 1 ) << i >> 1;
		ans += cnt[i] >> 1;
	}
	for ( int i = 1; i < 1 << 15; ++ i ) {
		for ( int j = 0; j < 15; ++ j ) {
			if ( ( i >> j ) & 1 ) {
				++ f[i], xsum[i] ^= j + 1;
			}
		}
		-- f[i];
	}
	for ( int s = 0; s < 1 << 15; ++ s ) {
		if ( xsum[s] ) continue;
		for ( int t = s; ; t = ( t - 1 ) & s ) {
			if ( ! xsum[t] && ! xsum[s ^ t] ) chkmin ( f[s], f[t] + f[s ^ t] );
			if ( ! t ) break;
		}
	}
	printf ( "%d\n", ans + f[S] );
	return 0;
}
```