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 INF = 0x7fffffff;
int a[2010];
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  int n, count = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] == 1) ++count;
  }
  if (count) {
    printf("%d", n - count);
  } else {
    int t = INF;
    int i, j;
    for (i = 1; i <= n; ++i) {
      int x = a[i];
      for (j = i + 1; j <= n; ++j) {
        x = gcd(x, a[j]);
        if (x == 1) {
          t = min(t, j - i + 1);
          break;
        }
      }
    }
    if (t == INF) {
      puts("-1");
    } else {
      printf("%d\n", t - 1 + n - 1);
    }
  }
  return 0;
}
