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>
const int maxn = 2010;
const int inf = 0x3f3f3f3f;
int a[maxn];
int min(int a, int b) { return a > b ? b : a; }
int gcd(int a, int b) { return a == 0 ? b : gcd(b % a, a); }
int main() {
  int n, l, r, i, j, tmp;
  while (~scanf("%d", &n)) {
    int cnt = 0;
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      if (a[i] == 1) cnt++;
    }
    if (cnt) {
      printf("%d\n", n - cnt);
      continue;
    }
    int mina = inf;
    for (i = 0; i < n; i++) {
      tmp = a[i];
      for (j = i + 1; j < n; j++) {
        tmp = gcd(tmp, a[j]);
        if (tmp == 1) {
          l = i;
          r = j;
          mina = min(mina, r - l);
          break;
        }
      }
    }
    if (mina != inf)
      printf("%d\n", mina + n - 1);
    else
      printf("-1\n");
  }
  return 0;
}
