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 a, int b) { return (a == 0) ? b : gcd(b % a, a); }
int in[2050];
int main() {
  int N, i, j;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) scanf("%d", &in[i]);
  int g = 0;
  for (i = 1; i <= N; i++) g = gcd(g, in[i]);
  if (g != 1) return !printf("-1\n");
  int c = 0;
  for (i = 1; i <= N; i++)
    if (in[i] == 1) c++;
  if (c) return !printf("%d\n", N - c);
  int mn = 0x3f3f3f3f;
  for (i = 1; i <= N; i++) {
    int g = in[i];
    for (j = i - 1; j >= 1; j--) {
      g = gcd(g, in[j]);
      if (g == 1) break;
    }
    if (j >= 1) mn = min(mn, i - j);
    g = in[i];
    for (j = i + 1; j <= N; j++) {
      g = gcd(g, in[j]);
      if (g == 1) break;
    }
    if (j <= N) mn = min(mn, j - i);
  }
  return !printf("%d\n", mn + N - 1);
}
