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 N = 2010;
int n, a[N], seg[4 * N];
inline int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
inline void build(int i, int lo, int hi) {
  if (lo == hi) {
    seg[i] = a[lo];
    return;
  }
  int mid = lo + hi >> 1;
  build(i << 1, lo, mid);
  build(i << 1 | 1, mid + 1, hi);
  seg[i] = gcd(seg[i << 1], seg[i << 1 | 1]);
}
inline int ask(int i, int lo, int hi, int ql, int qr) {
  if (qr < lo || hi < ql) return 0;
  if (ql <= lo && hi <= qr) return seg[i];
  int mid = lo + hi >> 1;
  return gcd(ask(i << 1, lo, mid, ql, qr),
             ask(i << 1 | 1, mid + 1, hi, ql, qr));
}
int main() {
  scanf("%d", &n);
  int cnt1 = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    if (a[i] == 1) cnt1++;
  }
  if (cnt1 > 0) {
    cout << n - cnt1 << endl;
    return 0;
  }
  build(1, 1, n);
  int lo = 0, hi = n + 1, mid;
  bool check;
  while (hi - lo > 1) {
    mid = lo + hi >> 1;
    check = false;
    for (int i = 1; i + mid - 1 <= n; i++)
      if (ask(1, 1, n, i, i + mid - 1) == 1) {
        check = 1;
        break;
      }
    if (check)
      hi = mid;
    else
      lo = mid;
  }
  if (hi == n + 1) {
    cout << -1;
    return 0;
  }
  cout << hi - 1 + (n - 1) << endl;
  return 0;
}
