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 n, cnt;
int p[2006];
int g;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    if (!i)
      g = p[i];
    else
      g = gcd(g, p[i]);
    if (p[i] == 1) cnt++;
  }
  int len = 0x3f3f3f3f;
  if (g > 1)
    printf("-1");
  else {
    if (cnt) {
      printf("%d\n", n - cnt);
    } else {
      for (int i = 0; i < n; i++) {
        int tg = p[i];
        for (int j = i + 1; j < n; j++) {
          tg = gcd(tg, p[j]);
          if (tg == 1) {
            len = min(len, j - i);
            break;
          }
        }
      }
      printf("%d\n", len + n - 1);
    }
  }
}
