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 gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  long long n, a[2009], g, cnt = 0;
  int len = INT_MAX;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i == 0)
      g = a[i];
    else
      g = gcd(g, a[i]);
    if (a[i] == 1) cnt++;
  }
  if (cnt > 0) {
    cout << (n - cnt) << endl;
    return 0;
  }
  if (g != 1) {
    cout << "-1" << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    g = a[i];
    for (int j = i + 1; j < n; j++) {
      g = gcd(g, a[j]);
      if (g == 1) len = min(len, (j - i + 1));
    }
  }
  cout << (n + len - 2) << endl;
  return 0;
}
