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;
long long a[2010];
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int n;
  scanf("%d", &n);
  int cnt = 0, ans = 0x3f3f3f3f;
  scanf("%lld", a + 1);
  if (a[1] == 1) cnt++;
  long long g = a[1];
  if (n == 1 && a[n] == 1) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 2; i <= n; i++) {
    scanf("%lld", a + i);
    if (a[i] == 1) cnt++;
    g = gcd(g, a[i]);
  }
  if (g != 1) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    long long temp = a[i];
    for (int j = i + 1; j <= n; j++) {
      int len;
      temp = gcd(temp, a[j]);
      if (temp == 1) {
        len = j - i;
        ans = min(len, ans);
        break;
      }
    }
    if (ans == 1) break;
  }
  printf("%d\n", ans + n - 1 - cnt);
}
