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;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  long long int a[n];
  long long int flag = 0;
  long long int s = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) s++;
  }
  for (long long int i = 0; i < n - 1; i++) {
    if (gcd(a[i], a[i + 1]) == 1) {
      flag = 1;
      break;
    }
  }
  if (s == n)
    cout << 0 << endl;
  else if (flag == 1)
    cout << n - s << endl;
  else {
    long long int mini = INT_MAX;
    for (long long int i = 0; i < n - 1; i++) {
      for (long long int j = i + 1; j < n - 1; j++) {
        long long int x = a[i];
        long long int y = a[j];
        long long int val = gcd(x, y);
        long long int c = 0;
        while (val != 1 && j < n - 1) {
          j++;
          val = gcd(val, a[j]);
          c++;
        }
        if (val == 1) mini = min(mini, c + 1);
        break;
      }
    }
    if (mini == INT_MAX)
      cout << -1 << endl;
    else
      cout << n - 1 + mini - s << endl;
  }
}
