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 b[2000];
int gcd(int a, int b) { return a == 0 ? b : gcd(b % a, a); }
int main() {
  int n;
  cin >> n;
  int num[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> num[i];
  }
  if (n == 1) {
    cout << (num[1] == 1 ? 0 : -1) << endl;
    return 0;
  }
  int flag = 0;
  int temp_min = 99999999;
  for (int i = 1; i <= n - 1; i++) {
    b[i] = gcd(num[i], num[i + 1]);
    if (b[i] == 1) {
      flag = 1;
      break;
    }
    if (b[i] < temp_min) {
      temp_min = b[i];
    }
  }
  if (flag) {
    int res = n;
    for (int i = 1; i <= n; i++) {
      if (num[i] == 1) {
        --res;
      }
    }
    cout << res << endl;
  } else {
    int count = 1;
    int flag2 = 0;
    int len = n - 1;
    while (count <= (n - 1)) {
      flag2 = 0;
      for (int i = 1; i <= len; i++) {
        b[i] = gcd(b[i], b[i + 1]);
        if (b[i] == 1) {
          cout << n + count << endl;
          return 0;
        }
      }
      --len;
      ++count;
    }
    cout << -1 << endl;
  }
  return 0;
}
