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 mod = 1e9 + 7;
const int MAXN = 2010;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int n;
vector<int> a;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    a.push_back(x);
  }
  if (n == 1 && a[0] != 1) {
    puts("-1");
    return 0;
  }
  int num = 0, cnt = 0;
  int ok = 0;
  for (int i = 0; i < a.size(); i++) {
    if (a[i] == 1) {
      ok++;
    }
  }
  if (ok) {
    printf("%d\n", n + max(cnt - 1, 0) - ok);
    return 0;
  }
  while (num < 2010) {
    num++;
    vector<int> ret;
    if (a[0] == 1) {
      printf("%d\n", n + max(0, cnt - 1));
      return 0;
    }
    for (int i = 1; i < a.size(); i++) {
      if (a[i] == 1) {
        printf("%d\n", n + max(0, cnt - 1));
        return 0;
      }
      ret.push_back(gcd(a[i], a[i - 1]));
    }
    if (ret.size() == 1 && ret[0] != 1) {
      puts("-1");
      return 0;
    }
    cnt++;
    a.clear();
    for (int i = 0; i < ret.size(); i++) {
      a.push_back(ret[i]);
    }
  }
  return 0;
}
