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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long a[n];
  long long ans = 0;
  long long cnt1 = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) cnt1++;
  }
  long long all = a[0];
  for (long long i = 0; i < n; i++) {
    all = gcd(all, a[i]);
  }
  if (all != 1) {
    cout << -1 << endl;
    return 0;
  }
  bool one_step = false;
  for (long long i = 0; i < n - 1; i++) {
    if (gcd(a[i], a[i + 1]) == 1) {
      cout << n - cnt1 << endl;
      return 0;
    }
  }
  ans = 1e17;
  for (long long i = 0; i < n; i++) {
    long long wet = a[i];
    for (long long j = i; j < n; j++) {
      wet = gcd(wet, a[j]);
      if (wet == 1) {
        ans = min(ans, n - 1 - j + 2 * (j - i) + i);
      }
    }
  }
  cout << ans << endl;
}
