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 a[100000];
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n, g, ans, i, j;
  scanf("%d", &n);
  ans = 1e9 + 7;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 1; i <= n; i++) {
    g = a[i];
    if (g == 1) {
      ans = 0;
      break;
    }
    for (j = i + 1; j <= n; j++) {
      g = gcd(a[j], g);
      if (g == 1) {
        ans = min(ans, j - i);
        break;
      }
    }
  }
  if (ans > 1e8)
    ans = -1;
  else if (ans == 0) {
    for (i = 1; i <= n; i++) {
      if (a[i] != 1) ans++;
    }
  } else
    ans = n + ans - 1;
  printf("%d\n", ans);
  return 0;
}
