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;
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
const double PI = acos(-1.0);
const long long mod = 1000000007;
const long long INF = 1e9;
const long long MAXN = 100001;
const long long MAXK = 20;
inline long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  long long G = a[0];
  for (int i = (1); i < (n); ++i) G = gcd(G, a[i]);
  if (G > 1) {
    cout << -1;
    return 0;
  }
  int ok = 0;
  for (int i = (0); i < (n); ++i)
    if (a[i] == 1) ++ok;
  if (ok) {
    cout << n - ok;
    return 0;
  }
  int length = n;
  for (int i = (0); i < (n); ++i) {
    G = a[i];
    for (int j = (i + 1); j < (n); ++j) {
      G = gcd(G, a[j]);
      if (G == 1) {
        length = min(length, j - i);
        break;
      }
    }
  }
  cout << n - 1 + length;
  return 0;
}
