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 arr[2003];
int len[2003];
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  int tot_gcd = arr[0];
  for (int i = 1; i < n; i++) {
    tot_gcd = (int)gcd(tot_gcd, arr[i]);
  }
  if (tot_gcd != 1) {
    cout << "-1\n";
    return 0;
  }
  int mn = 1e9;
  int satu = 0;
  for (int i = 0; i < n; i++) {
    int temp_gcd = arr[i];
    satu += 1 * (arr[i] == 1);
    for (int j = i; j < n; j++) {
      temp_gcd = gcd(temp_gcd, arr[j]);
      if (temp_gcd == 1) {
        mn = min(j - i + 1, mn);
        break;
      }
    }
  }
  if (satu) {
    printf("%d\n", n - satu);
    return 0;
  } else
    printf("%d\n", n + mn - 2);
}
