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) { return a ? gcd(b % a, a) : b; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<int> v(n);
  for (auto& it : v) cin >> it;
  int ones = 0;
  for (auto& it : v) ones += it == 1;
  if (ones) {
    cout << n - ones << endl;
    return 0;
  }
  int _gcd = v[0];
  for (auto& it : v) _gcd = gcd(_gcd, it);
  if (_gcd != 1) {
    cout << -1 << endl;
    return 0;
  }
  int closest_distance = n;
  for (int from = 0; from < n; from++) {
    _gcd = v[from];
    for (int to = from + 1; to < n; to++) {
      _gcd = gcd(_gcd, v[to]);
      if (_gcd == 1) {
        closest_distance = min(to - from - 1, closest_distance);
        break;
      }
    }
  }
  cout << closest_distance + n << endl;
}
