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;
int gcd1(int a, int b) {
  if (a == 0) return b;
  return gcd1(b % a, a);
}
long long modx(long long base, long long ex) {
  long long ans = 1LL, val = base;
  while (ex > 0LL) {
    if (ex & 1LL) ans = (ans * val) % 1000000009LL;
    val = (val * val) % 1000000009LL;
    ex = ex >> 1LL;
  }
  return ans;
}
int n;
int a[2005], g[2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    g[i] = gcd1(g[i - 1], a[i]);
    if (a[i] != 1) tot++;
  }
  if (tot == n) tot--;
  if (g[n] != 1) {
    cout << "-1" << endl;
    return 0;
  }
  int cnt = 2005;
  for (int i = 1; i <= n; i++) {
    if (g[i] == 1) {
      int x = a[i], tmp = 0;
      for (int j = i - 1; j >= 1 && x != 1; j--) {
        tmp++;
        x = gcd1(x, a[j]);
      }
      cnt = min(cnt, tmp);
    }
  }
  cout << cnt + tot << endl;
  return 0;
}
