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;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long int a[n + 1];
  for (int i = 1; i <= n; i++) cin >> a[i];
  int mini = n + 1;
  pair<int, int> p;
  for (int i = 1; i <= n; i++) {
    long long int k = a[i];
    for (int j = i; j <= n; j++) {
      k = gcd(k, a[j]);
      if (k == 1) {
        if (mini > j - i + 1) {
          mini = j - i + 1;
          p = make_pair(i, j);
        }
      }
    }
  }
  if (mini == n + 1) {
    cout << -1;
    return 0;
  }
  int ans = 0;
  for (int i = 2; i <= p.first; i++)
    if (a[i - 1] != 1) ans++;
  for (int i = p.second; i < n; i++)
    if (a[i + 1] != 1) ans++;
  ans += 2 * (p.second - p.first);
  cout << ans;
  return 0;
}
