#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
class LinearBase {
 public:
  const static int maxn = 64 - 1;
  long long d[maxn], p[maxn], tot;
  void clear() {
    tot = 0;
    memset(d, 0, sizeof(d));
    memset(p, 0, sizeof(p));
  }
  bool insert(long long val) {
    for (int i = maxn - 1; i >= 0; i--) {
      if ((val >> i) & 1) {
        if (!d[i]) {
          d[i] = val;
          break;
        } else {
          val ^= d[i];
        }
      }
    }
    return val > 0;
  }
  long long query_max(long long def = 0) {
    long long res = def;
    for (int i = maxn - 1; i >= 0; i--) {
      res = max(res, res ^ d[i]);
    }
    return res;
  }
  long long query_min() {
    for (int i = 0; i < maxn; i++) {
      if (d[i]) return d[i];
    }
    return 0;
  }
  void rebuild() {
    memset(p, 0, sizeof(p));
    tot = 0;
    for (int i = maxn - 1; i >= 0; i--) {
      for (int j = i - 1; j >= 0; j--) {
        if ((d[i] >> j) & 1) {
          d[i] ^= d[j];
        }
      }
    }
    for (int i = 0; i < maxn; i++) {
      if (d[i]) {
        p[tot++] = d[i];
      }
    }
  }
  long long query_kth(long long k) {
    if (k >= 1LL << tot) {
      return -1;
    }
    long long res = 0;
    for (int i = 0; i < maxn; i++) {
      if ((k >> i) & 1) {
        res ^= p[i];
      }
    }
    return res;
  }
} lb;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  while (cin >> n) {
    lb.clear();
    int now = 0;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      lb.insert(x);
      now ^= x;
    }
    lb.rebuild();
    cout << (now == 0 ? -1 : lb.tot) << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 maxn = 1e5 + 10;
class LinearBase {
 public:
  const static int maxn = 64 - 1;
  long long d[maxn], p[maxn], tot;
  void clear() {
    tot = 0;
    memset(d, 0, sizeof(d));
    memset(p, 0, sizeof(p));
  }
  bool insert(long long val) {
    for (int i = maxn - 1; i >= 0; i--) {
      if ((val >> i) & 1) {
        if (!d[i]) {
          d[i] = val;
          break;
        } else {
          val ^= d[i];
        }
      }
    }
    return val > 0;
  }
  long long query_max(long long def = 0) {
    long long res = def;
    for (int i = maxn - 1; i >= 0; i--) {
      res = max(res, res ^ d[i]);
    }
    return res;
  }
  long long query_min() {
    for (int i = 0; i < maxn; i++) {
      if (d[i]) return d[i];
    }
    return 0;
  }
  void rebuild() {
    memset(p, 0, sizeof(p));
    tot = 0;
    for (int i = maxn - 1; i >= 0; i--) {
      for (int j = i - 1; j >= 0; j--) {
        if ((d[i] >> j) & 1) {
          d[i] ^= d[j];
        }
      }
    }
    for (int i = 0; i < maxn; i++) {
      if (d[i]) {
        p[tot++] = d[i];
      }
    }
  }
  long long query_kth(long long k) {
    if (k >= 1LL << tot) {
      return -1;
    }
    long long res = 0;
    for (int i = 0; i < maxn; i++) {
      if ((k >> i) & 1) {
        res ^= p[i];
      }
    }
    return res;
  }
} lb;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  while (cin >> n) {
    lb.clear();
    int now = 0;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      lb.insert(x);
      now ^= x;
    }
    lb.rebuild();
    cout << (now == 0 ? -1 : lb.tot) << endl;
  }
  return 0;
}

```