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, a[2005];
int gcd(int a, int b) {
  while (b > 0) {
    int tmp = a % b;
    a = b;
    b = tmp;
  }
  return a;
}
void enter() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
}
void process() {
  int min_length = (int)2e9, L = 0, R = 0, cnt = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] == 1) cnt++;
  if (cnt > 0) {
    cout << n - cnt << endl;
    return;
  }
  for (int i = 1; i <= n; i++) {
    int igcd = a[i];
    if (igcd == 1) {
      min_length = 1;
      break;
    }
    for (int j = i + 1; j <= n; j++) {
      igcd = gcd(igcd, a[j]);
      if (igcd == 1) {
        min_length = min(min_length, j - i + 1);
        break;
      }
    }
  }
  if (min_length == (int)2e9) {
    cout << -1;
    return;
  }
  cout << min_length - 1 + (n - 1);
}
int main() {
  enter();
  process();
  return 0;
}
