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 vec[2000];
int gcd[2000][2000];
int min(int a, int b) { return a < b ? a : b; }
int Gcd(int a, int b) {
  if (b == 0) {
    return a;
  }
  return Gcd(b, a % b);
}
int Scm(int a, int b) { return 1LL * a * b / Gcd(a, b); }
int main() {
  int n;
  scanf("%d", &n);
  int res = n;
  int i;
  for (i = 0; i < n; ++i) {
    scanf("%d", &vec[i]);
    gcd[i][i] = vec[i];
    if (vec[i] <= 1) {
      res -= 1;
    }
  }
  if (res < n) {
    printf("%d\n", res);
    return 0;
  }
  int solution = 0;
  int len;
  for (len = 2; len <= n && !solution; ++len) {
    for (i = 0; i + len <= n && !solution; ++i) {
      gcd[i][i + len - 1] = Gcd(gcd[i][i], gcd[i + 1][i + len - 1]);
      if (gcd[i][i + len - 1] == 1) {
        solution = 1;
        res += len - 2;
      }
    }
  }
  printf("%d\n", (solution == 1 ? res : -1));
  return 0;
}
