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;
long long n, a[10000], mi = INT_MAX;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long dp[5000][5000];
long long check() {
  int cnt = 0;
  for (int ty = 0; ty < n; ty++) cnt += a[ty] == 1 ? 1 : 0;
  return cnt;
}
int main() {
  memset(dp, 0, sizeof(dp));
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long con = a[0];
  for (int i = 1; i < n; i++) {
    dp[i - 1][i] = gcd(a[i - 1], a[i]);
    if (dp[i - 1][i] == 1) mi = 2;
    con = gcd(con, a[i]);
  }
  if (con != 1)
    cout << -1 << endl;
  else if (n == 1 && a[0] != 1)
    cout << -1 << endl;
  else if (n == 1 && a[0] == 1)
    cout << 0 << endl;
  else {
    if (check() > 0)
      cout << n - check() << endl;
    else {
      for (int i = 0; i < n; i++) {
        for (int j = i + 2; j < n; j++) {
          dp[i][j] = gcd(dp[i][j - 1], a[j]);
          if (dp[i][j] == 1) mi = mi > (j - i + 1) ? j - i + 1 : mi;
        }
      }
      cout << mi + n - 2 << endl;
    }
  }
}
