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;
bool comp(pair<long long, long long> p1, pair<long long, long long> p2) {
  return p1.second > p2.second;
}
int gcd(int a, int b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<int> v(n);
  vector<vector<int> > sp(n, vector<int>(20));
  int b = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    sp[i][0] = v[i];
    if (v[i] == 1) b++;
  }
  if (n == 1) {
    if (v[0] == 1)
      cout << 0;
    else
      cout << -1;
    return 0;
  }
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 0; i < n; i++) {
      if (i + (1 << j) <= n) {
        sp[i][j] = gcd(sp[i][j - 1], sp[i + (1 << (j - 1))][j - 1]);
      } else
        sp[i][j] = sp[i][j - 1];
    }
  }
  int lo = 1;
  int hi = n - 1;
  bool k = 0;
  while (lo < hi) {
    int mid = (lo + hi) >> 1;
    int mi = mid;
    mid = mid + 1;
    bool ch = 0;
    for (int i = 0; i <= n - mid; i++) {
      int j = int(log2(mid));
      int h = gcd(sp[i][j], sp[i + mid - (1 << j)][j]);
      if (h == 1) {
        ch = 1;
        k = ch;
      }
    }
    if (ch)
      hi = mi;
    else
      lo = mi + 1;
  }
  for (int i = 0; i < n - lo; i++) {
    int j = int(log2(lo + 1));
    int h = gcd(sp[i][j], sp[i + lo + 1 - (1 << j)][j]);
    if (h == 1) {
      k = 1;
    }
  }
  if (k)
    cout << lo + n - 1 - b;
  else
    cout << -1;
  return 0;
}
