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>
const long long template_array_size = 1e6;
using namespace std;
long long gcd(long long a, long long b) {
  long long t;
  while (b != 0) {
    t = b;
    b = a % b;
    a = t;
  }
  return a;
}
long long lcm(long long a, long long b) {
  if (a > b)
    return (a / gcd(a, b)) * b;
  else
    return (b / gcd(a, b)) * a;
}
bool res = false, k = false;
void solve() {
  long long n, cnt = 0;
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    if (v[i] != 1) cnt++;
    if (v[i] == 1) k = true;
  }
  if (k) {
    cout << cnt;
    return;
  }
  long long ans = 1e17;
  for (long long i = 0; i < n; ++i) {
    long long g = 0;
    for (long long j = i; j < n; ++j) {
      g = gcd(g, v[j]);
      if (g == 1) ans = min(ans, j - i + 1);
    }
  }
  if (ans == 1e17)
    cout << -1;
  else
    cout << ans - 1 + n - 1;
}
int main() {
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
