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 NMAX = 2005;
const int INF = 0x7fffffff;
int v[NMAX];
int cmmdc(int a, int b) {
  while (b) {
    int r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  int n, nr1 = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &v[i]);
    nr1 += (v[i] == 1);
  }
  if (nr1 > 0) {
    printf("%d\n", n - nr1);
    return 0;
  }
  int sol = INF;
  for (int i = 1; i <= n; ++i) {
    int nr = v[i];
    for (int j = i + 1; j <= n && nr > 1; ++j) {
      nr = cmmdc(nr, v[j]);
      if (nr == 1) {
        sol = min(sol, j - i);
      }
    }
  }
  if (sol == INF) {
    printf("-1\n");
    return 0;
  }
  printf("%d\n", n - 1 + sol);
  return 0;
}
