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) {
  while (b) {
    int tmp = a;
    a = b;
    b = tmp % b;
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  int n;
  cin >> n;
  int one = 0;
  vector<int> vec(n), pref_gcd(n), suff_gcd(n);
  for (int i = 0; i < n; i++) {
    cin >> vec[i];
    if (vec[i] == 1) {
      one++;
    }
  }
  if (one) {
    cout << n - one;
    return 0;
  }
  int len = 1e9;
  for (int l = 0; l < n; l++) {
    int g = vec[l];
    for (int r = l + 1; r < n; r++) {
      g = gcd(g, vec[r]);
      if (g == 1) {
        len = min(len, r - l);
      }
    }
  }
  if (len == (int)1e9) {
    cout << -1;
    return 0;
  }
  cout << n + len - 1;
  return 0;
}
