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 maxx = 1e4 + 10;
int a[maxx];
int n;
int gcd(int x, int y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
int main() {
  while (scanf("%d", &n) != EOF) {
    int num = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      if (a[i] == 1) num++;
    }
    if (num) {
      printf("%d\n", n - num);
      continue;
    } else {
      int flag = 0;
      for (int i = 1; i <= n - 1; i++) {
        for (int j = 0; j < n - i; j++) {
          a[j] = gcd(a[j], a[j + 1]);
          if (a[j] == 1) {
            printf("%d\n", n + i - 1);
            flag = 1;
            break;
          }
        }
        if (flag) break;
      }
      if (!flag) printf("-1\n");
    }
  }
}
