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;
const int N = 2010;
int a[N], n, f[N][N];
inline int gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  scanf("%d", &n);
  int len = 200010, sum = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++)
    if (a[i] == 1) sum++;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      if (i == j)
        f[i][j] = a[i];
      else
        f[i][j] = gcd(f[i][j - 1], a[j]);
      if (f[i][j] == 1) {
        len = min(len, j - i + 1);
        break;
      }
    }
  }
  if (len == 200010) {
    printf("-1\n");
  } else {
    if (sum > 0) {
      printf("%d\n", n - sum);
    } else {
      int ans = len - 1 + n - 1;
      printf("%d\n", ans);
    }
  }
  return 0;
}
