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;
const int NMAX = 2000 + 100;
const int oo = 1e9;
int N;
int a[NMAX];
int gcd(int a, int b) {
  while (b != 0) {
    int r = a % b;
    a = b;
    b = r;
  }
  return a;
}
void Process() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &a[i]);
  int uc = a[1];
  for (int i = 2; i <= N; i++) uc = gcd(uc, a[i]);
  if (uc > 1) {
    printf("-1");
    return;
  }
  int cnt1 = 0;
  for (int i = 1; i <= N; i++)
    if (a[i] == 1) cnt1++;
  if (cnt1 > 0) {
    printf("%d", N - cnt1);
    return;
  }
  int res = oo;
  for (int i = 1; i <= N; i++) {
    int uc = a[i];
    for (int j = i; j <= N; j++) {
      uc = gcd(uc, a[j]);
      if (uc == 1) res = min(res, (j - i) + (N - 1));
    }
  }
  printf("%d", res);
}
int main() {
  Process();
  return 0;
}
