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;
const int mod = 1e9 + 7;
const int N = 2010;
int Gcd(int a, int b) { return b ? Gcd(b, a % b) : a; }
int a[N];
int g[N];
int main() {
  int n, ret = -1, num;
  cin >> n;
  int aa = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 1) aa++;
  }
  if (aa) {
    printf("%d\n", n - aa);
    return 0;
  }
  if (n == 1 && a[1] != 1) {
    puts("-1");
    return 0;
  } else if (n == 1 && a[1] == 1) {
    puts("1");
    return 0;
  }
  num = n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < num; j++) {
      g[j] = Gcd(a[j], a[j + 1]);
      a[j] = g[j];
      if (a[j] == 1) {
        ret = i;
        break;
      }
    }
    if (ret != -1) {
      ret = ret + n - 1;
      break;
    }
    num--;
  }
  cout << ret << endl;
  return 0;
}
