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 MOD = 1e7;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  int n, ans = INT_MAX, count = 0;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    count += (v[i] == 1);
  }
  if (count > 0) {
    cout << n - count << endl;
  } else {
    vector<vector<int>> dp(n, vector<int>(n));
    for (int i = 0; i < n; i++) dp[i][i] = v[i];
    for (int i = 0; i < n; i++)
      for (int j = i + 1; j < n; j++) dp[i][j] = gcd(v[j], dp[i][j - 1]);
    for (int i = 0; i < n; i++)
      for (int j = i; j < n; j++)
        if (dp[i][j] == 1) ans = min(ans, j - i);
    if (ans == INT_MAX)
      cout << -1 << endl;
    else
      cout << ans + n - 1 << endl;
  }
  return 0;
}
