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 x[2050] = {0};
int gcd(int a, int b) {
  if (a % b == 0) {
    return b;
  } else
    return gcd(b, a % b);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &x[i]);
  if (n == 1 && x[0] == 1) {
    printf("0\n");
    return 0;
  }
  int ans = 1000000;
  int ggg = 0;
  for (int j = 0; j < n - 1; j++) {
    int g = x[j], fin = 1, gg = 0;
    for (int i = j; i < n - 1; i++) {
      g = gcd(g, x[i + 1]);
      if (g == 1) {
        gg = 1;
        ggg = 1;
        break;
      }
      fin++;
    }
    if (gg) {
      if (fin < ans) ans = fin;
    }
  }
  if (ggg) {
    ans = ans + n - 1;
    for (int i = 0; i < n; i++)
      if (x[i] == 1) ans--;
    printf("%d\n", ans);
  } else
    printf("-1\n");
  return 0;
}
