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 b ? gcd(b, a % b) : a; }
signed main() {
  ios::sync_with_stdio(0);
  cin.sync_with_stdio(0);
  cin.tie(0);
  int N;
  cin >> N;
  int a[N + 1];
  bool h1 = false;
  for (int i = 1; i <= N; i++) {
    cin >> a[i];
    if (a[i] == 1) h1 = true;
  }
  if (h1) {
    int non = 0;
    for (int i = 1; i <= N; i++) {
      non += a[i] != 1;
    }
    cout << non << '\n';
    return 0;
  }
  int ans = INT_MAX;
  for (int i = 1; i <= N; i++) {
    int g = a[i];
    for (int j = i + 1; j <= N; j++) {
      g = gcd(g, a[j]);
      if (g == 1) {
        ans = min(ans, (j - i) + (N - 1));
      }
    }
  }
  if (ans == INT_MAX) ans = -1;
  cout << ans << '\n';
}
