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 num;
int arr[2010];
int gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    gcd(b, a % b);
}
int process(int val) {
  for (int i = 1; i <= num - val + 1; i++) {
    int ret = 0;
    for (int j = i; j <= i + val - 1; j++) {
      ret = gcd(ret, arr[j]);
    }
    if (ret == 1) return 1;
  }
  return 0;
}
int main(void) {
  scanf("%d", &num);
  for (int i = 1; i <= num; i++) scanf("%d", arr + i);
  int cnt = 0;
  for (int i = 1; i <= num; i++)
    if (arr[i] == 1) cnt++;
  if (cnt) {
    printf("%d", num - cnt);
    return 0;
  }
  int left = 1, right = num;
  while (left < right) {
    int mid = (left + right) / 2;
    if (process(mid))
      right = mid;
    else
      left = mid + 1;
  }
  if (process(left))
    printf("%d", left + num - 2);
  else
    printf("-1");
  return 0;
}
