#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
void exgcd(long long a, long long b, long long &d, long long &first,
           long long &second) {
  b                          ? exgcd(b, a % b, d, second, first),
      second -= a / b *first : first = 1, second = 0, d = a;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
const int N = 4e5 + 10, INF = 0x3f3f3f3f;
int n;
int a[N], b[N];
int main() {
  scanf("%d", &n);
  int sum = 0;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), sum ^= a[i];
  if (!sum) return puts("-1"), 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= *b; ++j) a[i] = min(a[i], a[i] ^ b[j]);
    if (a[i]) b[++*b] = a[i];
  }
  printf("%d\n", *b);
}

### Prompt

Generate a CPP solution to the following problem:

You are given an array a_1, a_2, ..., a_n of integer numbers.

Your task is to divide the array into the maximum number of segments in such a way that:

  * each element is contained in exactly one segment; 
  * each segment contains at least one element; 
  * there doesn't exist a non-empty subset of segments such that bitwise XOR of the numbers from them is equal to 0. 



Print the maximum number of segments the array can be divided into. Print -1 if no suitable division exists.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the array.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).

Output

Print the maximum number of segments the array can be divided into while following the given constraints. Print -1 if no suitable division exists.

Examples

Input


4
5 5 7 2


Output


2


Input


3
1 2 3


Output


-1


Input


3
3 1 10


Output


3

Note

In the first example 2 is the maximum number. If you divide the array into \{[5], [5, 7, 2]\}, the XOR value of the subset of only the second segment is 5 ⊕ 7 ⊕ 2 = 0. \{[5, 5], [7, 2]\} has the value of the subset of only the first segment being 5 ⊕ 5 = 0. However, \{[5, 5, 7], [2]\} will lead to subsets \{[5, 5, 7]\} of XOR 7, \{[2]\} of XOR 2 and \{[5, 5, 7], [2]\} of XOR 5 ⊕ 5 ⊕ 7 ⊕ 2 = 5.

Let's take a look at some division on 3 segments — \{[5], [5, 7], [2]\}. It will produce subsets:

  * \{[5]\}, XOR 5; 
  * \{[5, 7]\}, XOR 2; 
  * \{[5], [5, 7]\}, XOR 7; 
  * \{[2]\}, XOR 2; 
  * \{[5], [2]\}, XOR 7; 
  * \{[5, 7], [2]\}, XOR 0; 
  * \{[5], [5, 7], [2]\}, XOR 5; 



As you can see, subset \{[5, 7], [2]\} has its XOR equal to 0, which is unacceptable. You can check that for other divisions of size 3 or 4, non-empty subset with 0 XOR always exists.

The second example has no suitable divisions.

The third example array can be divided into \{[3], [1], [10]\}. No subset of these segments has its XOR equal to 0.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
void exgcd(long long a, long long b, long long &d, long long &first,
           long long &second) {
  b                          ? exgcd(b, a % b, d, second, first),
      second -= a / b *first : first = 1, second = 0, d = a;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
const int N = 4e5 + 10, INF = 0x3f3f3f3f;
int n;
int a[N], b[N];
int main() {
  scanf("%d", &n);
  int sum = 0;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), sum ^= a[i];
  if (!sum) return puts("-1"), 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= *b; ++j) a[i] = min(a[i], a[i] ^ b[j]);
    if (a[i]) b[++*b] = a[i];
  }
  printf("%d\n", *b);
}

```