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 N = 2e5 + 5;
const int INF = 1e9;
const long long mod = 1e9 + 7;
int n, arr[N], byk;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    if (arr[i] == 1) byk++;
  }
  if (byk) {
    cout << n - byk << endl;
    return 0;
  }
  int mini = INF;
  for (int i = 1; i < n; i++) {
    int now = gcd(arr[i], arr[i + 1]);
    int cnt = 1;
    for (int j = i; j >= 1; j--) {
      if (now == 1) {
        mini = min(mini, cnt);
        break;
      }
      now = gcd(now, arr[j - 1]);
      cnt++;
    }
    now = gcd(arr[i], arr[i + 1]);
    cnt = 1;
    for (int j = i + 1; j <= n; j++) {
      if (now == 1) {
        mini = min(mini, cnt);
        break;
      }
      now = gcd(now, arr[j + 1]);
      cnt++;
    }
  }
  if (mini == INF) {
    cout << -1 << endl;
    return 0;
  }
  cout << mini + (n - 1) << endl;
}
