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;
template <class T>
inline void read(T &x) {
  int f = 0, ch = 0;
  x = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = 1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  if (f) x = -x;
}
int n;
long long a[3005];
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main() {
  read(n);
  int ans = (0x7f7f7f7f) / 3, num = n;
  for (int i = 1; i <= n; i++) {
    read(a[i]);
    if (a[i] == 1) num--;
  }
  for (int i = 1; i <= n; i++) {
    int now = a[i];
    if (a[i] == 1) {
      ans = 1;
      break;
    }
    for (int j = i + 1; j <= n; j++) {
      now = gcd(now, a[j]);
      if (now == 1) {
        ans = ((ans) < (j - i) ? (ans) : (j - i));
        break;
      }
    }
  }
  if (ans != (0x7f7f7f7f) / 3)
    printf("%d", ans - 1 + num);
  else
    printf("-1");
  return 0;
}
