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 a, int b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  int n, d, ones = 0;
  cin >> n;
  int A[n];
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    if (A[i] == 1) ones++;
  }
  d = A[0];
  for (int i = 1; i < n && d > 1; i++) d = gcd(A[i], d);
  if (d > 1)
    cout << -1 << endl;
  else if (ones > 0)
    cout << n - ones << endl;
  else {
    int ans = n, d, j;
    for (int i = 0; i < n - 1; i++) {
      d = A[i], j = i + 1;
      for (; j < n && d > 1; j++) d = gcd(d, A[j]);
      if (d == 1) ans = min(ans, j - i - 1);
    }
    cout << (n - 1) + ans << endl;
  }
  return 0;
}
