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 MX = 3e5 + 1, MOD = 1e9 + 7;
int a[MX], b[MX], n, k, h, c, m, x, y, z, ans = 0;
int f[4];
int gcd(int x, int y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  x = a[0];
  int pos = 0;
  for (int i = 1; i < n; i++) {
    x = gcd(x, a[i]);
  }
  if (x > 1)
    cout << "-1\n";
  else {
    int cnt = 0;
    for (int i = 0; i < n; i++)
      if (a[i] == 1) cnt++;
    if (cnt > 0)
      cout << n - cnt << "\n";
    else {
      int ans = n + 1;
      for (int i = 0; i < n; i++) {
        int y = a[i];
        for (int j = i; j < n; j++) {
          y = gcd(y, a[j]);
          if (y == 1) {
            ans = min(ans, j - i);
            break;
          }
        }
      }
      cout << n - 1 + ans;
    }
  }
}
int32_t main() {
  int t = 1;
  while (t--) {
    solve();
  }
}
