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 gi() {
  int a;
  scanf("%d", &a);
  return a;
}
long long gli() {
  long long a;
  scanf("%I64d", &a);
  return a;
}
int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a > b) return gcd(a % b, b);
  return gcd(b % a, a);
}
int a[2000];
int main() {
  int g = 0;
  int n = gi();
  int n1 = 0;
  for (int i = 0; i < n; i++) {
    a[i] = gi();
    if (a[i] == 1) {
      g = 1;
      n1++;
    } else {
      g = gcd(g, a[i]);
    }
  }
  if (g > 1) {
    printf("-1\n");
    return 0;
  }
  if (n1 > 0) {
    printf("%d\n", n - n1);
    return 0;
  }
  int res = n;
  while (1) {
    for (int i = 0; i < n - 1; i++) {
      a[i] = gcd(a[i], a[i + 1]);
      if (a[i] == 1) {
        printf("%d\n", res);
        return 0;
      }
    }
    n--;
    res++;
  }
  return 0;
}
