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;
vector<int> A;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int findGCD(vector<int> arr, int n) {
  int result = arr[0];
  for (int i = 1; i < n; i++) result = gcd(arr[i], result);
  return result;
}
int main() {
  int n;
  cin >> n;
  A.resize(n);
  for (int i = 0; i < n; i++) cin >> A[i];
  int apgd = findGCD(A, n);
  if (apgd != 1) {
    cout << "-1" << endl;
    return 0;
  }
  int b1 = 0;
  int e1 = n - 1;
  for (int i = 0; i < n; i++) {
    if (A[i] == 1) {
      int bb = 0;
      for (int i = 0; i < n; i++)
        if (A[i] != 1) bb++;
      cout << bb << endl;
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int res = A[i];
    for (int j = i + 1; j < n; j++) {
      res = gcd(res, A[j]);
      if (res == 1) {
        if (j - i + 1 < e1 - b1 + 1) {
          b1 = i;
          e1 = j;
          break;
        }
      }
      if (j - i + 1 >= e1 - b1 + 1) break;
    }
  }
  cout << e1 - b1 + n - 1 << endl;
  return 0;
}
