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 N, av[2003];
void readin() {
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    cin >> av[i];
  }
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
void process() {
  int flag = 0;
  for (int i = 1; i <= N; ++i) {
    if (av[i] == 1) {
      flag++;
    }
  }
  if (flag) {
    cout << N - flag << endl;
    return;
  }
  flag = 1;
  int ans = INT_MAX;
  for (int i = 1; i <= N; ++i) {
    int t = av[i];
    for (int j = i + 1; j <= N; ++j) {
      if ((t = gcd(t, av[j])) == 1) {
        ans = min(ans, j - i + N - 1);
        break;
      }
    }
  }
  if (ans == INT_MAX) {
    cout << -1 << endl;
  } else {
    cout << ans << endl;
  }
}
int main() {
  readin();
  process();
  return 0;
}
