You have an array a with length n, you can perform operations. Each operation is like this: choose two adjacent elements from a, say x and y, and replace one of them with gcd(x, y), where gcd denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).

What is the minimum number of operations you need to make all of the elements equal to 1?

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2000) — the number of elements in the array.

The second line contains n space separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1.

Examples

Input

5
2 2 3 4 6


Output

5


Input

4
2 4 6 8


Output

-1


Input

3
2 6 9


Output

4

Note

In the first sample you can turn all numbers to 1 using the following 5 moves:

  * [2, 2, 3, 4, 6]. 
  * [2, 1, 3, 4, 6]
  * [2, 1, 3, 1, 6]
  * [2, 1, 1, 1, 6]
  * [1, 1, 1, 1, 6]
  * [1, 1, 1, 1, 1]



We can prove that in this case it is not possible to make all numbers one using less than 5 moves.
#include <bits/stdc++.h>
using namespace std;
int n, arr[2002], same, cnt, found, temp;
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  while (count(arr, arr + n, 1) != n) {
    same = 1;
    for (int i = 0; i < n - 1; ++i) {
      if (arr[i] != arr[i + 1]) same = 0;
      if (gcd(arr[i], arr[i + 1]) == 1 && arr[i] != arr[i + 1]) {
        ++cnt;
        found = 1;
        if (arr[i] != 1)
          arr[i] = 1;
        else
          arr[i + 1] = 1;
      }
    }
    if (same) return cout << -1, 0;
    if (!same && !found) {
      for (int i = 0; i < n - 1; ++i) {
        if (arr[i] != arr[i + 1]) {
          ++cnt;
          temp = gcd(arr[i], arr[i + 1]);
          if (i != n - 1 && gcd(temp, arr[i + 2]) == 1)
            arr[i + 1] = temp;
          else if (arr[i] > arr[i + 1] && arr[i] != temp)
            arr[i] = temp;
          else
            arr[i + 1] = temp;
        }
      }
    }
  }
  cout << cnt;
  return 0;
}
