#include <bits/stdc++.h>
using namespace std;
const int N = 2000010, mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long a[N], b[N];
int main() {
  int n = read();
  for (int i = 0; i < n; ++i) a[i] = read(), b[i] = b[i - 1] ^ a[i];
  if (b[n - 1] == 0) {
    puts("-1");
    return 0;
  }
  int k = 0;
  for (int i = 31; i >= 0; --i) {
    for (int j = k; j < n; ++j) {
      if (b[j] >> i & 1) {
        swap(b[k], b[j]);
        break;
      }
    }
    if (!(b[k] >> i & 1)) continue;
    for (int j = 0; j < n; ++j) {
      if (j != k && (b[j] >> i & 1)) b[j] ^= b[k];
    }
    k++;
    if (k == n) break;
  }
  cout << k;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 N = 2000010, mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long a[N], b[N];
int main() {
  int n = read();
  for (int i = 0; i < n; ++i) a[i] = read(), b[i] = b[i - 1] ^ a[i];
  if (b[n - 1] == 0) {
    puts("-1");
    return 0;
  }
  int k = 0;
  for (int i = 31; i >= 0; --i) {
    for (int j = k; j < n; ++j) {
      if (b[j] >> i & 1) {
        swap(b[k], b[j]);
        break;
      }
    }
    if (!(b[k] >> i & 1)) continue;
    for (int j = 0; j < n; ++j) {
      if (j != k && (b[j] >> i & 1)) b[j] ^= b[k];
    }
    k++;
    if (k == n) break;
  }
  cout << k;
  return 0;
}

```