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 a[20000], n, cnt, x, y;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) cnt++;
  }
  if (cnt > 0) {
    cout << n - cnt;
    return 0;
  }
  cnt = 0;
  y = 1;
  if (n == 1)
    if (a[0] == 1) {
      cout << 0;
      return 0;
    } else {
      cout << -1;
      return 0;
    }
  bool f = 0;
  while (gcd(a[x], a[y]) != 1) {
    if (x == n - 2 && y == n - 1) {
      f = 1;
      break;
    }
    x++;
    y++;
  }
  x = 0;
  y = 1;
  if (f) {
    while (a[x] == a[y]) {
      x++;
      y++;
    }
    while (gcd(a[x], a[y]) != 1 && y != n) {
      cnt++;
      a[y] = gcd(a[x], a[y]);
      x++;
      y++;
    }
    if (y == n) {
      cout << -1;
      return 0;
    }
  } else {
    cout << n;
    return 0;
  }
  cout << n + cnt;
  return 0;
}
