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) {
  long long c;
  while (b != 0) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
int n;
vector<long long> a;
int main() {
  cin >> n;
  a.resize(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long g1 = a[0];
  long long i1 = 0;
  for (long long i : a) {
    g1 = gcd(i, g1);
    if (i == 1) i1++;
  }
  if (g1 > 1) {
    cout << -1 << endl;
    return 0;
  }
  if (i1 > 0) {
    cout << n - i1 << endl;
    return 0;
  }
  long long mini = n * n;
  for (long long i = 0; i < n; i++) {
    long long g = a[i];
    for (long long j = i + 1; j < n; j++) {
      g = gcd(a[j], g);
      if (g == 1) {
        mini = min(mini, j - i + n - 1);
      }
    }
  }
  cout << mini << endl;
  return 0;
}
