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 NMAX = 2001;
int v[NMAX];
int cmmdc(int a, int b) {
  int r = a % b;
  while (r) {
    a = b;
    b = r;
    r = a % b;
  }
  return b;
}
int main() {
  int n, i, k, ok, flag;
  scanf("%d", &n);
  scanf("%d", &v[1]);
  ok = v[1];
  flag = (v[1] == 1);
  for (i = 2; i <= n; i++) {
    scanf("%d", &v[i]);
    if (v[i] == 1) flag++;
    ok = cmmdc(ok, v[i]);
  }
  if (ok > 1)
    printf("-1\n");
  else if (flag == 0) {
    k = 2, ok = 0;
    while (!ok) {
      for (i = 1; i <= n - k + 1; i++) {
        v[i] = cmmdc(v[i], v[i + 1]);
        if (v[i] == 1) ok = 1;
      }
      k++;
    }
    k--;
    printf("%d\n", n + k - 2);
  } else
    printf("%d\n", n - flag);
  return 0;
}
