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;
ll gcd(ll u, ll v) {
  while (v != 0) {
    ll r = u % v;
    u = v;
    v = r;
  }
  return u;
}
int getBest(vector<ll>& vec, int start, int best) {
  ll curGcd = vec[start];
  for (int i = start + 1; i < vec.size(); ++i) {
    curGcd = gcd(curGcd, vec[i]);
    if (curGcd == 1) {
      return i - start;
    }
  }
  return 3000;
}
int main() {
  int n;
  cin >> n;
  int one_count = 0;
  vector<ll> vec(n);
  for (int i = 0; i < n; ++i) {
    ll num;
    cin >> num;
    if (num == 1) {
      one_count++;
    }
    vec[i] = num;
  }
  if (one_count > 0) {
    cout << n - one_count << endl;
    return 0;
  }
  int best = 3000;
  for (int i = 0; i < n - 1; ++i) {
    int _count = getBest(vec, i, best);
    best = min(best, _count);
  }
  if (best == 3000) {
    cout << "-1" << endl;
    return 0;
  }
  cout << n + best - 1 << endl;
  return 0;
}
