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;
  gcd(b, a % b);
}
long long int A[2100];
int main() {
  int n;
  cin >> n;
  int flag = -1;
  long long int s;
  long long int count1 = 0;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    if (i != 0) {
      s = gcd(A[i], A[i - 1]);
      if (s == 1) flag = 1;
    }
    if (A[i] == 1) {
      count1++;
      flag = 1;
    }
  }
  if (flag == 1) {
    cout << n - count1 << endl;
  } else {
    long long int minL = INT_MAX;
    for (int i = 0; i < n; i++) {
      int g = A[i];
      for (int j = i + 1; j < n; j++) {
        g = gcd(A[j], g);
        if (g == 1) {
          int len = j - i + 1;
          if (len < minL) minL = len;
        }
      }
    }
    if (minL >= INT_MAX)
      cout << "-1" << endl;
    else
      cout << (minL - 2 + n) << endl;
  }
}
