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 z = 0;
  while (y > 0) {
    z = y;
    y = x % y;
    x = z;
  }
  return x;
}
int main() {
  int n;
  cin >> n;
  vector<int> a;
  int ed = 0;
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    a.push_back(t);
    if (t == 1) {
      ed += 1;
    }
  }
  if (ed > 0) {
    cout << n - ed;
  } else {
    vector<int> mas;
    int ind = 0;
    for (int l = 0; l < n; l++) {
      int s = a[l];
      for (int r = l + 1; r < n; r++) {
        s = gcd(s, a[r]);
        if (s == 1) {
          ind = 1;
          mas.push_back(r - l);
        }
      }
    }
    if (ind == 0) {
      cout << -1;
    } else {
      int minx = mas[0];
      for (int i = 1; i < mas.size(); i++) {
        if (mas[i] < minx) {
          minx = mas[i];
        }
      }
      cout << n - 1 + minx;
    }
  }
}
