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.
/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class TaskC {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();

    int n = Integer.parseInt(br.readLine());
    int[] vals = new int[n];
    boolean has1 = false;
    String[] s = br.readLine().split(" ");
    int runningGcd = Integer.parseInt(s[0]);
    for (int i=0; i<n; i++) {
      vals[i] = Integer.parseInt(s[i]);
      if (vals[i]==1) {
        has1 = true;
      }
      runningGcd = gcd(runningGcd, vals[i]);
    }
    if (has1) {
      int non1 = 0;
      for (int i=0; i<n; i++) {
        if (vals[i]!=1) {
          non1++;
        }
      }
      System.out.println(non1);
      return;
    }
    if (runningGcd!=1) {
      System.out.println("-1");
      return;
    }
    int minRun = n;
    for (int i=0; i<n; i++) {
      runningGcd = vals[i];
      for (int j=1; j<minRun && i+j<n; j++) {
        runningGcd = gcd(runningGcd, vals[i+j]);
        if (runningGcd==1) {
          minRun = j+1;
        }
      }
    }
    System.out.println(minRun-1+(n-1));

  }
  static int gcd(int a, int b) {
    return (b==0) ? a : gcd(b, a%b);
  }
}
