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 gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int n, a[10010], sum, l = 1, r, cnt = 0;
bool check(int x) {
  int tem;
  for (int i = 1; i <= n - x + 1; i++) {
    tem = a[i];
    for (int j = i + 1; j < i + x; j++) {
      tem = gcd(tem, a[j]);
    }
    if (tem == 1) return true;
  }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cnt++;
    if (i == 1)
      sum = a[i];
    else
      sum = gcd(sum, a[i]);
  }
  if (sum != 1) {
    puts("-1");
    return 0;
  }
  r = n;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid)) {
      r = mid - 1;
    } else
      l = mid + 1;
  }
  if (cnt == 0)
    printf("%d\n", l + n - 2);
  else
    printf("%d\n", n - cnt);
}
