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 isPrime(long long k) {
  if (k <= 1) {
    return 0;
  }
  if (k == 2) {
    return 1;
  }
  if (k % 2 == 0) {
    return 0;
  }
  long long i = 3;
  while (i * i <= k) {
    if (k % i == 0) {
      return 0;
    }
    i += 2;
  }
  return 1;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long g = 0, l;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) g++;
  }
  if (g > 0) {
    cout << n - g;
  } else {
    l = n + 1;
    for (long long i = 0; i < n - 1; i++) {
      g = a[i];
      for (long long h = i + 1; h < n; h++) {
        g = gcd(g, a[h]);
        if (g == 1) {
          l = min(l, h - i + 1);
          break;
        }
      }
    }
    if (l != n + 1) {
      cout << l + n - 2;
    } else {
      cout << -1;
    }
  }
  return 0;
}
