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 int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
const int INF = INT_MAX;
const long long INFLL = (long long)1e18;
const double PI = acos(-1.0);
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
const int N = (int)1e5 + 5;
const int P = 1 << 18;
int n;
int a[2005];
int main() {
  ios::sync_with_stdio(false);
  cout.setf(ios::fixed | ios::showpoint);
  cout.precision(8);
  cin >> n;
  int cur;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i == 0) {
      cur = a[i];
    } else
      cur = gcd(cur, a[i]);
  }
  if (cur != 1) return !printf("-1\n");
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] != 1) cnt++;
  }
  if (cnt != n) {
    return !printf("%d\n", cnt);
  }
  int ans = INT_MAX;
  for (int i = 0; i < n; i++) {
    int cur = a[i];
    for (int j = i + 1; j < n; j++) {
      if (gcd(cur, a[j]) == 1) {
        ans = min(ans, j - i + n - 1);
      }
    }
  }
  if (ans == INT_MAX) ans = n + 1;
  cout << ans << endl;
  return 0;
}
