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>
#pragma warning(disable : 4996)
using namespace std;
int n, m, i, j, k, l, r, x;
int a[2009], b[2009];
int ones;
int gcd(int u, int v) {
  while (v != 0) {
    int r = u % v;
    u = v;
    v = r;
  }
  return u;
}
int solve(int start) {
  int res = 0;
  for (int i = start + 1; i < n && ones < n; i++) {
    b[i] = gcd(b[i], b[i - 1]);
    if (b[i] == 1) ones++;
    res++;
  }
  for (int i = n - 2; i >= 0 && ones < n; i--) {
    b[i] = gcd(b[i], b[i + 1]);
    if (b[i] == 1) ones++;
    res++;
  }
  for (int i = 0; i < n && ones < n; i++) {
    b[i] = gcd(b[i], b[i - 1]);
    if (b[i] == 1) ones++;
    res++;
  }
  if (ones < n) return INT_MAX;
  return res;
}
int main() {
  cin >> n;
  int minres = INT_MAX;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < n; i++) {
    ones = 0;
    for (j = 0; j < n; j++) {
      b[j] = a[j];
      if (b[j] == 1) ones++;
    }
    minres = min(solve(i), minres);
  }
  if (minres == INT_MAX)
    cout << -1;
  else
    cout << minres;
  return 0;
}
