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;
void speed() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  long long int test = -1, oneCount = 0;
  bool flag = false;
  for (long long int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] == 1) ++oneCount;
    if (i != 0) {
      if (test == -1) {
        test = gcd(a[0], a[i]);
      } else {
        test = gcd(test, a[i]);
      }
      if (flag == false && gcd(a[i], a[i - 1]) == 1) flag = true;
    }
  }
  long long int k = 0;
  for (; k < n; ++k) {
    if (a[k] != 1) break;
  }
  if (k == n) {
    cout << 0;
    return;
  }
  if (test != 1) {
    cout << -1;
    return;
  }
  if (flag == 1) {
    cout << n - oneCount;
    return;
  }
  long long int ans = 0;
  for (long long int i = 1; i < n; ++i) {
    if (a[i] != a[i - 1]) {
      a[i] = gcd(a[i], a[i - 1]);
      ++ans;
    }
    if (a[i] == 1) {
      break;
    }
  }
  for (long long int i = 0; i < n; ++i) {
    if (a[i] != 1) ++ans;
  }
  cout << ans;
}
int main() { solve(); }
