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 MAXN = 2000 + 5;
int sequence[MAXN];
int pre[MAXN];
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int n;
  scanf("%d", &n);
  int count = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &sequence[i]);
    if (sequence[i] == 1) {
      count++;
    }
  }
  for (int i = 1; i < n; i++) {
    pre[i] = gcd(sequence[i], sequence[i - 1]);
  }
  int min_distance = MAXN;
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (gcd(pre[i], sequence[j]) == 1) {
        if (min_distance > j - i) {
          min_distance = j - i;
        }
      }
    }
  }
  if (min_distance == MAXN) {
    if (n == 1 && sequence[0] == 1) {
      printf("0\n");
    } else {
      printf("-1\n");
    }
  } else {
    int result = min_distance - count + n;
    printf("%d\n", result);
  }
  return 0;
}
