#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char s) { return string(1, s); }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = 1;
  string r = "{";
  for (const auto &x : v) {
    if (!f) r += ", ";
    f = 0;
    r += to_string(x);
  }
  return r + "}";
}
void debug_out() { cout << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << " " << to_string(H);
  debug_out(T...);
}
inline int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
inline int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
inline int mul(int a, int b) { return (int)((long long)a * b % MOD); }
inline int binpow(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b /= 2;
  }
  return res;
}
inline int inv(int a) { return binpow(a, MOD - 2); }
int gcd(int a, int b, int &x, int &y) {
  if (a == 0) {
    x = 0, y = 1;
    return b;
  }
  int x1, y1;
  int d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  int all = 0;
  for (int &x : a) {
    cin >> x;
    all ^= x;
  }
  if (!all) {
    cout << "-1\n";
    return 0;
  }
  const int LOG = 30;
  int sz = 0;
  for (int bit = LOG - 1; bit >= 0 && sz < n; --bit) {
    for (int i = sz; i < n; ++i)
      if (a[i] & (1 << bit)) {
        swap(a[i], a[sz]);
        break;
      }
    if (!(a[sz] & (1 << bit))) continue;
    for (int i = sz + 1; i < n; ++i)
      if (a[i] & (1 << bit)) a[i] ^= a[sz];
    sz++;
  }
  cout << sz << "\n";
  return 0;
}

### Prompt

Generate a Cpp solution to 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 MOD = 1e9 + 7;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char s) { return string(1, s); }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = 1;
  string r = "{";
  for (const auto &x : v) {
    if (!f) r += ", ";
    f = 0;
    r += to_string(x);
  }
  return r + "}";
}
void debug_out() { cout << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << " " << to_string(H);
  debug_out(T...);
}
inline int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
inline int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
inline int mul(int a, int b) { return (int)((long long)a * b % MOD); }
inline int binpow(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b /= 2;
  }
  return res;
}
inline int inv(int a) { return binpow(a, MOD - 2); }
int gcd(int a, int b, int &x, int &y) {
  if (a == 0) {
    x = 0, y = 1;
    return b;
  }
  int x1, y1;
  int d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  int all = 0;
  for (int &x : a) {
    cin >> x;
    all ^= x;
  }
  if (!all) {
    cout << "-1\n";
    return 0;
  }
  const int LOG = 30;
  int sz = 0;
  for (int bit = LOG - 1; bit >= 0 && sz < n; --bit) {
    for (int i = sz; i < n; ++i)
      if (a[i] & (1 << bit)) {
        swap(a[i], a[sz]);
        break;
      }
    if (!(a[sz] & (1 << bit))) continue;
    for (int i = sz + 1; i < n; ++i)
      if (a[i] & (1 << bit)) a[i] ^= a[sz];
    sz++;
  }
  cout << sz << "\n";
  return 0;
}

```