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;
int a[1000010];
int gcd(int a, int b) {
  int r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  int f = gcd(a[1], a[2]);
  for (int i = 3; i <= n; ++i) {
    f = gcd(f, a[i]);
  }
  if (f != 1) {
    cout << "-1";
    return 0;
  }
  int nr = 0, i, j, min = 0;
  for (i = 1; i <= n; ++i)
    if (a[i] != 1) ++nr;
  if (nr != n) {
    cout << nr;
    return 0;
  }
  for (i = 1; i < n; ++i) {
    int mut = 1;
    f = gcd(a[i], a[i + 1]);
    for (j = i + 2; f != 1 && j <= n; ++j) {
      ++mut;
      f = gcd(f, a[j]);
    }
    if (min == 0 || (mut < min && f == 1)) min = mut;
  }
  cout << min + n - 1;
  return 0;
}
