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 gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
int dix[] = {1, -1, 0, 0, 1, 1, -1, -1};
int diy[] = {0, 0, 1, -1, 1, -1, 1, -1};
long long SP[2222][100];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<long long> v(n);
  int one = 00, gc;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    one += bool(v[i] == 1);
  }
  gc = v[0];
  for (int i = 1; i < n; i++) gc = gcd(v[i], gc);
  if (one) {
    cout << n - one << '\n';
    return 0;
  }
  int lgk = log2(n);
  for (int i = 0; i < n; i++) SP[i][0] = v[i];
  for (int j = 1; j <= lgk; j++) {
    for (int i = 0; (i + (1 << j) - 1) < n; i++) {
      SP[i][j] = gcd(SP[i][j - 1], SP[i + (1 << (j - 1))][j - 1]);
    }
  }
  int res = 1e8;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int k = log2(j - i + 1);
      if ((gcd(SP[i][k], SP[j - (1 << k) + 1][k])) == 1)
        res = min(res, n - 1 + j - i);
    }
  }
  if (res == 1e8) res = -1;
  cout << res << '\n';
  return 0;
  ;
}
