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 int MAXN = (1e5) + 10;
const long long int MINN = 1e2;
const long long int inf = 1e9 + 10;
int gcd(int a, int b) {
  if (a < b) swap(a, b);
  if (b)
    return gcd(b, a % b);
  else
    return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin.exceptions(cin.failbit);
  int n, cnt = 0;
  vector<int> v;
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    int x;
    cin >> x;
    v.push_back(x);
  }
  long long x = 0;
  for (int i = 0; i < (n); ++i)
    if (v[i] == 1) ++x;
  if (x) {
    cout << n - x << endl;
    exit(0);
  }
  while (1) {
    for (int i = 0; i < (v.size()); ++i) {
      if (v[i] == 1) {
        cout << cnt + n - 1 << endl;
        exit(0);
      }
    }
    if (v.size() == 1) break;
    ++cnt;
    vector<int> tmp;
    for (int i = 0; i < (v.size() - 1); ++i) tmp.push_back(gcd(v[i], v[i + 1]));
    v.clear();
    for (int i = 0; i < (tmp.size()); ++i) v.push_back(tmp[i]);
  }
  cout << "-1" << endl;
  return 0;
}
