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 MAX = 40000000;
int n;
vector<int> v;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void check() {
  int qtd = 0;
  for (int i = 0; i < n; i++) {
    if (v[i] == 1) qtd++;
  }
  if (qtd) {
    printf("%d\n", n - qtd);
    exit(0);
  }
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    v.push_back(x);
  }
  check();
  int mini = 10000000;
  for (int i = 0; i < n; i++) {
    int g = v[i];
    for (int j = i + 1; j < n; j++) {
      g = gcd(g, v[j]);
      if (g == 1) {
        mini = min(j - i + 1, mini);
      }
    }
  }
  if (mini == 10000000)
    puts("-1");
  else
    printf("%d\n", mini + n - 2);
  return 0;
}
