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 a[2050];
int GCD(int x, int y) { return x % y == 0 ? y : GCD(y, x % y); }
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  int L = 0;
  int Count = N;
  int i;
  int flag = 0;
  for (i = 1; i <= N; i++) {
    cin >> a[i];
    if (a[i] == 1) {
      flag++;
    }
  }
  if (flag) {
    cout << N - flag;
    exit(0);
  }
  while (Count != 1) {
    L++;
    for (i = 1; i < Count; i++) {
      a[i] = GCD(a[i], a[i + 1]);
      if (a[i] == 1) {
        flag = 1;
        break;
      }
    }
    if (flag) break;
    Count--;
  }
  if (flag) {
    cout << N - 1 + L << endl;
  } else
    cout << -1 << endl;
  return 0;
}
