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 n, m, ar[1000005], sum, flag, mn = INT_MAX;
long long int gcd(long long int a, long long int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> ar[i];
  for (int i = 1; i <= n; i++)
    if (ar[i] == 1) flag++;
  if (flag) return cout << n - flag, 0;
  for (int i = 1; i <= n; i++) {
    long long int x = 0;
    for (int j = i; j <= n; j++) {
      x = gcd(x, ar[j]);
      if (x == 1) mn = min(mn, (long long int)j - i);
    }
  }
  if (mn == INT_MAX) return cout << -1 << "\n", 0;
  cout << n + mn - 1;
}
