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;
  return gcd(b, a % b);
}
int main() {
  long long int n, i;
  cin >> n;
  long long int a[n], cnt = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) cnt++;
  }
  if (n == 1) {
    cout << (a[0] == 1 ? 0 : -1) << endl;
    return 0;
  }
  if (cnt) {
    cout << n - cnt << endl;
    return 0;
  }
  long long int ans = 9000000000000000000;
  for (long long int i = 0; i < n; i++) {
    long long int cnt1 = 0, cnt2 = 0;
    long long int gd1 = a[i], gd2 = a[i];
    for (long long int j = i + 1; j < n; j++) {
      cnt1++;
      gd1 = gcd(gd1, a[j]);
      if (gd1 == 1) break;
    }
    for (long long int j = i - 1; j >= 0; j--) {
      cnt2++;
      gd2 = gcd(gd2, a[j]);
      if (gd2 == 1) break;
    }
    if (gd1 == 1) ans = min(ans, cnt1 + n - 1);
    if (gd2 == 1) ans = min(ans, cnt2 + n - 1);
  }
  cout << (ans == 9000000000000000000 ? -1 : ans) << endl;
  cerr << "time taken : " << (float)clock() / (float)CLOCKS_PER_SEC << " secs"
       << endl;
  return 0;
}
