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 gcd(int x, int y) {
  if (x == y) return x;
  while (x != y) {
    if (x > y)
      x = x - y;
    else
      y = y - x;
  }
  return x;
}
int minx(int x, int y) {
  if (x == y) return x;
  if (x > y)
    return y;
  else
    return x;
}
int main() {
  int arr[2000];
  int ds = 2001;
  int n;
  int check1 = 0;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  for (int i = 0; i < n; ++i) {
    if (arr[i] == 1) {
      check1++;
    }
  }
  if (check1 != 0) {
    cout << n - check1;
    return 0;
  }
  for (int i = 0; i < n - 1; ++i) {
    if (arr[i] == arr[i + 1]) continue;
    int x = arr[i];
    for (int j = i + 1; j < n; ++j) {
      x = gcd(x, arr[j]);
      if (x == 1) {
        ds = minx(ds, j - i);
        break;
      }
    }
  }
  if (ds != 2001)
    cout << n - 1 + ds;
  else
    cout << "-1";
  return 0;
}
