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 n;
long long A[2005];
long long gcd(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
  }
  int rangeSize = 100000;
  for (int i = 1; i <= n; i++) {
    long long rangeGCD = A[i];
    for (int j = i; j <= n; j++) {
      rangeGCD = gcd(rangeGCD, A[j]);
      if (rangeGCD == 1) {
        rangeSize = min(rangeSize, j - i + 1);
      }
    }
  }
  if (rangeSize == 100000) {
    cout << -1 << '\n';
  } else {
    int cnt_one = 0;
    for (int i = 1; i <= n; i++) {
      if (A[i] == 1) cnt_one++;
    }
    int minimum;
    if (cnt_one == 0)
      minimum = rangeSize - 1 + n - 1;
    else
      minimum = n - cnt_one;
    cout << minimum << '\n';
  }
  return 0;
}
