#include <bits/stdc++.h>
using namespace std;
char ch;
bool fh;
inline void read(int &a) {
  for (fh = 0, ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') fh = 1;
  for (a = 0; ch >= '0' && ch <= '9'; ch = getchar()) (a *= 10) += ch - '0';
  if (fh) a = -a;
}
int A[200010];
int n, ans;
int r[32];
void linearsubset(int len) {
  int vi, vj, tmp;
  for (vi = n; vi >= 1; vi--) {
    tmp = A[vi];
    for (vj = len - 1; vj >= 0; vj--)
      if ((tmp >> vj) & 1) {
        if (r[vj])
          tmp ^= r[vj];
        else {
          r[vj] = tmp;
          ans++;
          break;
        }
      }
  }
}
int main() {
  int vi, maxx = 0;
  read(n);
  for (vi = 1; vi <= n; vi++) read(A[vi]);
  for (vi = 1; vi <= n; vi++) A[vi] ^= A[vi - 1], maxx = max(maxx, A[vi]);
  int leng = floor(log2(maxx) + 1);
  if (!A[n])
    printf("-1");
  else {
    linearsubset(leng);
    printf("%d\n", ans);
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
char ch;
bool fh;
inline void read(int &a) {
  for (fh = 0, ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') fh = 1;
  for (a = 0; ch >= '0' && ch <= '9'; ch = getchar()) (a *= 10) += ch - '0';
  if (fh) a = -a;
}
int A[200010];
int n, ans;
int r[32];
void linearsubset(int len) {
  int vi, vj, tmp;
  for (vi = n; vi >= 1; vi--) {
    tmp = A[vi];
    for (vj = len - 1; vj >= 0; vj--)
      if ((tmp >> vj) & 1) {
        if (r[vj])
          tmp ^= r[vj];
        else {
          r[vj] = tmp;
          ans++;
          break;
        }
      }
  }
}
int main() {
  int vi, maxx = 0;
  read(n);
  for (vi = 1; vi <= n; vi++) read(A[vi]);
  for (vi = 1; vi <= n; vi++) A[vi] ^= A[vi - 1], maxx = max(maxx, A[vi]);
  int leng = floor(log2(maxx) + 1);
  if (!A[n])
    printf("-1");
  else {
    linearsubset(leng);
    printf("%d\n", ans);
  }
}

```