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 x, int y) {
  int i = 2, ans = 1;
  while (x != 1) {
    if (x % i == 0) {
      if (y % i == 0) {
        ans *= i, y /= i;
      }
      x /= i;
    } else {
      i++;
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, cnt_one = 0;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] == 1) cnt_one++;
  }
  if (0 < cnt_one) {
    cout << n - cnt_one;
  } else {
    int ans = INT_MAX;
    for (int l = 0; l < n; l++) {
      int temp_gcd = a[l];
      for (int r = l + 1; r < n; r++) {
        temp_gcd = gcd(temp_gcd, a[r]);
        if (temp_gcd == 1) {
          ans = min(r - l, ans);
          break;
        }
      }
    }
    if (ans != INT_MAX) {
      cout << ans + n - 1;
    } else {
      cout << -1;
    }
  }
  return 0;
}
