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>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
int a, b, c, d, n, m, k;
int mas[2001];
inline int gcd(int a, int b) {
  for (; b; swap(a, b)) a %= b;
  return a;
}
int main() {
  scanf("%d", &n);
  a = 0;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    scanf("%d", &mas[i]);
    a = gcd(a, mas[i]);
  }
  if (a > 1) {
    printf("-1\n");
    exit(0);
  }
  a = 0;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    if (mas[i] != 1) ++a;
  }
  c = 1000000000;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    if (mas[i] == 1) {
      c = 0;
      break;
    }
    int cur = 0, d = mas[i];
    for (int _n(n - 1), j(i + 1); j <= _n; j++) {
      ++cur;
      d = gcd(d, mas[j]);
      if (d == 1) break;
    }
    if (d == 1) c = min(c, cur - 1);
  }
  printf("%d\n", a + c);
}
