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 GCC optimize "-O3"
using namespace std;
long long gcd(long long a, long long b) {
  while (b) {
    long long q = a % b;
    a = b;
    b = q;
  }
  return a;
}
const int MAXN = 2100;
int a[MAXN];
int n;
const int INF = 1e6;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  int cc1 = 0;
  for (int i = 0; i < n; ++i)
    if (a[i] == 1) {
      ++cc1;
    }
  if (cc1) {
    cout << n - cc1 << "\n";
    return 0;
  }
  int ans = INF;
  for (int i = 0; i < n; ++i) {
    int cur = a[i];
    for (int j = i - 1; j >= 0; --j) {
      cur = gcd(cur, a[j]);
      if (cur == 1) {
        ans = min(ans, n + i - j - 1);
        break;
      }
    }
    cur = a[i];
    for (int j = i + 1; j < n; ++j) {
      cur = gcd(cur, a[j]);
      if (cur == 1) {
        ans = min(ans, n + j - i - 1);
        break;
      }
    }
  }
  if (ans == INF)
    cout << -1 << "\n";
  else
    cout << ans << "\n";
  return 0;
}
