#include <bits/stdc++.h>
using namespace std;
struct linear_base {
  int a[66], mxlog;
  void Insert(int x) {
    for (int i = mxlog; i >= 0; i--) {
      if (x & (1ll << i)) {
        if (!a[i]) {
          a[i] = x;
          return;
        } else
          x ^= a[i];
      }
    }
  }
  bool check(int x) {
    for (int i = mxlog; i >= 0; i--) {
      if (x & (1ll << i)) {
        if (!a[i])
          return 0;
        else
          x ^= a[i];
      }
    }
    return 1;
  }
  void merge(const linear_base &b) {
    for (int i = mxlog; i >= 0; i--) {
      if (b.a[i]) Insert(b.a[i]);
    }
  }
  int findmax() {
    int res = 0;
    for (int i = mxlog; i >= 0; i--) {
      if ((res ^ a[i]) > res) res ^= a[i];
    }
    return res;
  }
  int findmin() {
    for (int i = 0; i <= mxlog; i++) {
      if (a[i]) return a[i];
    }
  }
  int findkth(int k) {
    int tmp[66], cnt = 0, res = 0;
    for (int i = 0; i <= mxlog; i++) {
      for (int j = i - 1; j >= 0; j--) {
        if (a[i] & (1ll << j)) a[i] ^= a[j];
      }
      if (a[i]) tmp[cnt++] = a[i];
    }
    for (int i = 0; i < cnt; i++) {
      if (k & (1ll << i)) res ^= tmp[i];
    }
    return res;
  }
} l;
int n, a[222222], pre[222222], ans, cnt;
int main() {
  srand(time(0));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] ^ a[i];
  l.mxlog = 30;
  if (!pre[n]) {
    printf("-1\n");
    return 0;
  }
  for (int t = 1; t <= 50; t++) {
    random_shuffle(pre + 1, pre + n);
    memset(l.a, 0, sizeof(l.a));
    cnt = 1;
    l.Insert(pre[n]);
    for (int i = 1; i < n; i++) {
      if (l.check(pre[i]) || !pre[i]) continue;
      l.Insert(pre[i]);
      cnt++;
    }
    ans = max(ans, cnt);
  }
  printf("%d\n", ans);
  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;
struct linear_base {
  int a[66], mxlog;
  void Insert(int x) {
    for (int i = mxlog; i >= 0; i--) {
      if (x & (1ll << i)) {
        if (!a[i]) {
          a[i] = x;
          return;
        } else
          x ^= a[i];
      }
    }
  }
  bool check(int x) {
    for (int i = mxlog; i >= 0; i--) {
      if (x & (1ll << i)) {
        if (!a[i])
          return 0;
        else
          x ^= a[i];
      }
    }
    return 1;
  }
  void merge(const linear_base &b) {
    for (int i = mxlog; i >= 0; i--) {
      if (b.a[i]) Insert(b.a[i]);
    }
  }
  int findmax() {
    int res = 0;
    for (int i = mxlog; i >= 0; i--) {
      if ((res ^ a[i]) > res) res ^= a[i];
    }
    return res;
  }
  int findmin() {
    for (int i = 0; i <= mxlog; i++) {
      if (a[i]) return a[i];
    }
  }
  int findkth(int k) {
    int tmp[66], cnt = 0, res = 0;
    for (int i = 0; i <= mxlog; i++) {
      for (int j = i - 1; j >= 0; j--) {
        if (a[i] & (1ll << j)) a[i] ^= a[j];
      }
      if (a[i]) tmp[cnt++] = a[i];
    }
    for (int i = 0; i < cnt; i++) {
      if (k & (1ll << i)) res ^= tmp[i];
    }
    return res;
  }
} l;
int n, a[222222], pre[222222], ans, cnt;
int main() {
  srand(time(0));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] ^ a[i];
  l.mxlog = 30;
  if (!pre[n]) {
    printf("-1\n");
    return 0;
  }
  for (int t = 1; t <= 50; t++) {
    random_shuffle(pre + 1, pre + n);
    memset(l.a, 0, sizeof(l.a));
    cnt = 1;
    l.Insert(pre[n]);
    for (int i = 1; i < n; i++) {
      if (l.check(pre[i]) || !pre[i]) continue;
      l.Insert(pre[i]);
      cnt++;
    }
    ans = max(ans, cnt);
  }
  printf("%d\n", ans);
  return 0;
}

```