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 read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long n;
long long a[2000 + 5];
void clean() {}
void solve() {
  clean();
  long long onen = 0;
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    if (a[i] == 1) onen++;
  }
  if (n == 1) {
    if (a[1] == 1) {
      printf("0\n");
      return;
    } else {
      printf("-1\n");
      return;
    }
  }
  long long ans = 1000000000 + 5;
  for (long long woc = 1; woc < n; woc++) {
    long long g = gcd(a[woc], a[woc + 1]);
    long long whw = 0;
    if (g == 1) {
      onen++, whw = woc;
    } else
      for (long long i = woc + 2; i <= n; i++) {
        long long tmp2 = gcd(a[i], g);
        if (tmp2 == 1) {
          onen++, whw = i - 1, g = tmp2;
          break;
        }
      }
    if (g == 1)
      ans = min(ans, n - onen + whw);
    else if (woc == 1) {
      printf("-1\n");
      return;
    }
  }
  printf("%I64d\n", ans);
}
int main() {
  scanf("%I64d", &n), solve();
  return 0;
}
