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[2005];
int gcd(int a, int b) {
  int temp;
  while (b) {
    temp = b;
    b = a % b;
    a = temp;
  }
  return a;
}
int main() {
  int n, sum = 0, now;
  scanf("%d", &n);
  scanf("%d", &a[1]);
  now = a[1];
  if (a[1] == 1) sum++;
  for (int i = 2; i <= n; i++) {
    scanf("%d", &a[i]);
    now = gcd(now, a[i]);
    if (a[i] == 1) {
      sum++;
    }
  }
  if (now != 1) {
    printf("-1\n");
  } else if (sum != 0) {
    int ans = n - sum;
    printf("%d\n", ans);
  } else {
    int q, k, o = 9999999;
    for (int i = 1; i <= n; i++) {
      q = a[i];
      k = 0;
      for (int j = i + 1; j <= n; j++) {
        q = gcd(q, a[j]);
        k++;
        if (q == 1) {
          if (o > k) o = k;
          break;
        }
      }
    }
    int ans = o + n - 1;
    printf("%d\n", ans);
  }
  return 0;
}
