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 double pi = acos(-1);
const double eps = 1e6 - 9;
const int inf = (1 << 30);
const int N = 2005;
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int dp[N][N];
int a[N];
int n;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cout.precision(15);
  cin >> n;
  for (int i = 1; i <= n; i += +1) {
    cin >> a[i];
    dp[i][i] = a[i];
  }
  for (int l = 1; l <= n; l += +1) {
    for (int r = l + 1; r <= n; r += +1) {
      dp[l][r] = gcd(dp[l][r - 1], a[r]);
    }
  }
  for (int replace = 1; replace <= n; replace += +1) {
    bool became_one = false;
    for (int l = 1; l <= n - replace + 1; l += +1) {
      int r = l + replace - 1;
      if (dp[l][r] == 1) {
        became_one = true;
        break;
      }
    }
    if (became_one) {
      if (replace == 1) {
        int ones = count_if(a, a + n + 1, [=](int x) { return x == 1; });
        cout << n - ones << '\n';
      } else {
        cout << n + replace - 2 << '\n';
      }
      return 0;
    }
  }
  cout << "-1\n";
  return 0;
}
