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>
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n, count_one = 0, isdealt = 0;
  long long int num[2005];
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%I64d", &num[i]);
  for (int i = 0; i < n; i++)
    if (num[i] == 1) {
      isdealt = 1;
      count_one++;
    }
  if (isdealt) {
    printf("%d", n - count_one);
    return 0;
  }
  for (int i = 1; i <= n - 1; i++)
    for (int j = 0; j < n - i; j++) {
      num[j] = gcd(num[j], num[j + 1]);
      if (num[j] == 1) {
        printf("%d", i + n - 1);
        return 0;
      }
    }
  printf("-1");
  return 0;
}
