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[100010];
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1, n, i, j;
  cin >> n;
  int cnt = 0;
  for (i = 0; i < n; i++) {
    cin >> A[i];
    if (A[i] == 1) cnt++;
  }
  if (cnt) {
    cout << n - cnt << "\n";
  } else {
    int cur = 0;
    for (i = 0; i < n; i++) cur = gcd(cur, A[i]);
    if (cur != 1) {
      cout << "-1\n";
    } else {
      int mn = n;
      for (i = 0; i < n; i++) {
        cur = A[i];
        for (int j = i + 1; j < n; j++) {
          cur = gcd(cur, A[j]);
          if (cur == 1) {
            mn = min(mn, j - i);
            break;
          }
        }
      }
      cout << mn + n - 1 << "\n";
    }
  }
  return 0;
  return 0;
}
