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 maxn = 1e6 + 5;
const int ff = 0x3f3f3f3f;
int n;
int a[5200];
int gcd(int x, int y) {
  if (x < y) x = x ^ y, y = x ^ y, x = x ^ y;
  return x % y == 0 ? y : gcd(y, x % y);
}
int main() {
  cin >> n;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cnt++;
  }
  if (cnt) {
    cout << n - cnt << endl;
    return 0;
  }
  int mint = ff;
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    cnt = 0;
    for (int j = i + 1; j <= n; j++) {
      cnt++;
      x = gcd(x, a[j]);
      if (x == 1) break;
    }
    if (x == 1) mint = min(mint, cnt);
  }
  if (mint == ff)
    cout << -1 << endl;
  else
    cout << mint + n - 1 << endl;
  return 0;
}
