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>
int gcd(int a, int b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
int minimum(int a, int b) { return (a < b) ? a : b; }
int pride(int *arr, int n) {
  int ones = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] == 1) ones++;
  }
  if (ones > 0) return n - ones;
  int min = INT_MAX;
  for (int i = 0; i < n; i++) {
    int a = arr[i];
    for (int j = i + 1; j < n; j++) {
      a = gcd(a, arr[j]);
      if (a == 1) {
        min = minimum(min, j - i);
        break;
      }
    }
  }
  return (min == INT_MAX) ? -1 : min + n - 1;
}
int main(void) {
  int n;
  scanf("%d", &n);
  int arr[n];
  for (int i = 0; i < n; i++) {
    int temp;
    scanf("%d", &temp);
    arr[i] = temp;
  }
  printf("%d", pride(arr, n));
}
