#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <class S, class T>
inline S smax(S &l, T r) {
  return l = max(l, (S)r);
}
template <class S, class T>
inline S smin(S &l, T r) {
  return l = min(l, (S)r);
}
const long long MAXN = 2e5 + 10;
const long long L = 31;
long long n, a[MAXN], p[MAXN], basis[MAXN];
void transform(long long x) {
  for (long long i = L - 1; ~i; --i)
    if (basis[i] != -1 && (x & (1 << i))) x ^= basis[i];
  if (x == 0) return;
  for (long long i = L - 1; ~i; --i)
    if (x & (1 << i)) {
      basis[i] = x;
      return;
    }
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  memset(basis, -1, sizeof(basis));
  for (long long i = 0; i < n; ++i) {
    p[i + 1] = p[i] ^ a[i];
    transform(p[i + 1]);
  }
  if (p[n] == 0) {
    cout << -1 << '\n';
    return 0;
  }
  long long cnt = 0;
  for (long long i = 0; i < L; ++i)
    if (basis[i] != -1) cnt++;
  cout << cnt << '\n';
  return 0;
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <class S, class T>
inline S smax(S &l, T r) {
  return l = max(l, (S)r);
}
template <class S, class T>
inline S smin(S &l, T r) {
  return l = min(l, (S)r);
}
const long long MAXN = 2e5 + 10;
const long long L = 31;
long long n, a[MAXN], p[MAXN], basis[MAXN];
void transform(long long x) {
  for (long long i = L - 1; ~i; --i)
    if (basis[i] != -1 && (x & (1 << i))) x ^= basis[i];
  if (x == 0) return;
  for (long long i = L - 1; ~i; --i)
    if (x & (1 << i)) {
      basis[i] = x;
      return;
    }
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  memset(basis, -1, sizeof(basis));
  for (long long i = 0; i < n; ++i) {
    p[i + 1] = p[i] ^ a[i];
    transform(p[i + 1]);
  }
  if (p[n] == 0) {
    cout << -1 << '\n';
    return 0;
  }
  long long cnt = 0;
  for (long long i = 0; i < L; ++i)
    if (basis[i] != -1) cnt++;
  cout << cnt << '\n';
  return 0;
}

```