#include <bits/stdc++.h>
using namespace std;
const long long INF = INT_MAX;
const long long MOD = 1e9 + 7;
const int MAX_ROW = 40;
const int MAX_COL = 200010;
struct BitMatrix {
  int H, W;
  bitset<MAX_COL> val[MAX_ROW];
  BitMatrix(int m = 1, int n = 1) : H(m), W(n) {}
  inline bitset<MAX_COL>& operator[](int i) { return val[i]; }
};
int GaussJordan(BitMatrix& A, bool is_extended = false) {
  int rank = 0;
  for (int col = 0; col < A.W; ++col) {
    if (is_extended && col == A.W - 1) break;
    int pivot = -1;
    for (int row = rank; row < A.H; ++row) {
      if (A[row][col]) {
        pivot = row;
        break;
      }
    }
    if (pivot == -1) continue;
    swap(A[pivot], A[rank]);
    for (int row = 0; row < A.H; ++row) {
      if (row != rank && A[row][col]) A[row] ^= A[rank];
    }
    ++rank;
  }
  return rank;
}
int main() {
  int n;
  cin >> n;
  int x = 0;
  BitMatrix Mat(40, n);
  for (int i = 0; (i) < (n); ++(i)) {
    int a;
    cin >> a;
    x ^= a;
    for (int j = 0; (j) < (40); ++(j)) {
      Mat[j][i] = (a >> j) & 1;
    }
  }
  int rank = GaussJordan(Mat);
  cout << (x ? rank : -1) << endl;
  return 0;
}

### 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;
const long long INF = INT_MAX;
const long long MOD = 1e9 + 7;
const int MAX_ROW = 40;
const int MAX_COL = 200010;
struct BitMatrix {
  int H, W;
  bitset<MAX_COL> val[MAX_ROW];
  BitMatrix(int m = 1, int n = 1) : H(m), W(n) {}
  inline bitset<MAX_COL>& operator[](int i) { return val[i]; }
};
int GaussJordan(BitMatrix& A, bool is_extended = false) {
  int rank = 0;
  for (int col = 0; col < A.W; ++col) {
    if (is_extended && col == A.W - 1) break;
    int pivot = -1;
    for (int row = rank; row < A.H; ++row) {
      if (A[row][col]) {
        pivot = row;
        break;
      }
    }
    if (pivot == -1) continue;
    swap(A[pivot], A[rank]);
    for (int row = 0; row < A.H; ++row) {
      if (row != rank && A[row][col]) A[row] ^= A[rank];
    }
    ++rank;
  }
  return rank;
}
int main() {
  int n;
  cin >> n;
  int x = 0;
  BitMatrix Mat(40, n);
  for (int i = 0; (i) < (n); ++(i)) {
    int a;
    cin >> a;
    x ^= a;
    for (int j = 0; (j) < (40); ++(j)) {
      Mat[j][i] = (a >> j) & 1;
    }
  }
  int rank = GaussJordan(Mat);
  cout << (x ? rank : -1) << endl;
  return 0;
}

```