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;
void mohammed() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main() {
  mohammed();
  int n, ones = 0;
  long long arr[2001];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 1) ones++;
  }
  if (ones) {
    cout << n - ones;
    return 0;
  }
  int a = n;
  bool z = false;
  for (int i = 0; i < n; i++) {
    long long g = arr[i];
    int j = i + 1, c = 0;
    while (j < n) {
      g = gcd(g, arr[j]);
      c++;
      if (g == 1) {
        z = true;
        if (c < a) {
          a = c;
        }
      }
      j++;
    }
  }
  if (z) {
    cout << n + a - 1;
  } else
    cout << -1;
  return 0;
}
