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;
const long long INF = 1e9 + 7;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void solve() {
  long long n;
  cin >> n;
  long long arr[n];
  for (long long i = 0; i < (n); i++) cin >> arr[i];
  long long g = arr[0];
  for (long long i = 0; i < (n); i++) g = gcd(g, arr[i]);
  if (g != 1)
    cout << "-1" << endl;
  else {
    long long mn = INF;
    long long cnt = 0;
    for (long long i = 0; i < (n); i++)
      if (arr[i] == 1) cnt++;
    if (cnt) {
      cout << n - cnt << endl;
      return;
    }
    for (long long i = 0; i < n; i++) {
      g = arr[i];
      for (long long j = i; j >= 0; j--) {
        g = gcd(g, arr[j]);
        if (g == 1) {
          mn = min(mn, i - j + 1);
          break;
        }
      }
    }
    cout << n + mn - 2 << endl;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) solve();
}
