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) {
  if (!a) return b;
  return gcd(b % a, a);
}
int main() {
  int n;
  cin >> n;
  int a[n], count = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] == 1) count++;
  }
  int diff = 1e5, f = 0, temp;
  if (!count) {
    for (int i = 0; i < n; ++i) {
      temp = a[i];
      for (int j = i; j < n; j++) {
        temp = gcd(temp, a[j]);
        if (temp == 1) {
          f = 1;
          if (j - i < diff) diff = j - i;
        }
      }
    }
    if (!f)
      cout << -1 << endl;
    else {
      cout << n + diff - 1 << endl;
    }
  } else {
    cout << n - count << endl;
  }
}
