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 mygcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return mygcd(b, a % b);
}
int main() {
  int n, k, temp, c;
  int count = 0;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 1) count++;
  }
  if (count > 0) {
    cout << n - count << endl;
    return 0;
  }
  vector<int> dp(n, -1);
  for (int j = 0; j < n; j++) {
    k = j - 1;
    temp = arr[j];
    c = 0;
    while (k >= 0 and temp != 1) {
      temp = mygcd(temp, arr[k]);
      c++;
      k--;
    }
    if (temp == 1) dp[j] = c;
  }
  int mins = INT_MAX;
  for (int j = 0; j < n; j++) {
    if (dp[j] != -1) mins = min(mins, dp[j]);
  }
  if (mins == INT_MAX)
    cout << -1 << endl;
  else
    cout << n - 1 + mins << endl;
}
