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;
using ll = long long;
const int N = 2e5 + 100;
const int INF = 2e9;
int t[N];
int gcd(int a, int b) { return (a == 0) ? b : gcd(b % a, a); }
bool czy(int n, int length) {
  for (int i = 0; i <= n - length; i++) {
    int res = t[i];
    for (int j = i + 1; j < i + length; j++) {
      res = gcd(res, t[j]);
    }
    if (res == 1) {
      return true;
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int ile = 0;
  for (int i = 0; i < n; i++) {
    cin >> t[i];
    if (t[i] == 1) ile++;
  }
  if (ile) {
    cout << n - ile << "\n";
    return 0;
  }
  int l = 2;
  int r = n + 1;
  while (l < r) {
    int s = (l + r) / 2;
    if (czy(n, s)) {
      r = s;
    } else {
      l = s + 1;
    }
  }
  if (l == n + 1) {
    cout << "-1"
         << "\n";
    return 0;
  }
  cout << n + l - 2 << "\n";
}
