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;
using LL = long long;
void Print() { cout << endl; }
template <typename T1, typename... T>
void Print(const T1 &t1, const T &...t) {
  cout << t1 << " ";
  Print(t...);
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <class Iterator>
ostream &output_cantainer(ostream &os, Iterator first, Iterator last,
                          bool space = true) {
  if (space) {
    for (Iterator it = first; it != last; it++) {
      os << (*it) << ' ';
    }
    os << endl;
  } else
    for (Iterator it = first; it != last; it++) {
      os << (*it) << endl;
    }
  return os;
}
const int N = 2003, INF = 10000;
int n, a[N], dp[N];
int gcd(int a, int b) {
  while (b) {
    int r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  dp[1] = (a[1] == 1) ? 0 : INF;
  for (int i = 2; i <= n; i++) {
    dp[i] = min(INF, dp[i - 1] + 1);
    int j = i, v = a[i];
    for (j--; j >= 1 && v > 1; j--) {
      v = gcd(v, a[j]);
    }
    if (v == 1) {
      j++;
      int s = i - j;
      for (; j >= 1; j--) {
        dp[i] = min(dp[i], i - j + s + dp[j - 1]);
      }
    }
  }
  printf("%d\n", (dp[n] == INF) ? -1 : dp[n]);
}
