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 arr[3000];
int min_ = 99999999;
int gcd(int m, int n) {
  if (n < m) {
    int c = n;
    n = m;
    m = c;
  }
  while (m) {
    int c = n % m;
    n = m;
    m = c;
  }
  return n;
}
int flag = 0, eq1flag = 0;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) eq1flag++;
  }
  if (eq1flag) {
    printf("%d\n", n - eq1flag);
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int t = 0; t < n - i - 1; t++) {
      int tmp = gcd(arr[t], arr[t + 1]);
      arr[t] = tmp;
      if (tmp == 1) {
        printf("%d\n", n + i);
        return 0;
      }
    }
  }
  printf("-1\n");
  return 0;
}
