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 sortbyfir(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.second == b.second) return a.first < b.first;
  return (a.second < b.second);
}
bool sr(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first)
    return a.second > b.second;
  else
    return a.first < b.first;
}
bool compmx(int a, int b) { return (a < b); }
long long int binaryToDecimal(string n) {
  string num = n;
  long long int dec_value = 0;
  long long int base = 1;
  for (int i = num.length() - 1; i >= 0; i--) {
    if (num[i] == '1') dec_value += base;
    base = base * 2;
  }
  return dec_value;
}
string decimalToBinary(long long int n) {
  string s = bitset<64>(n).to_string();
  const auto loc1 = s.find('1');
  if (loc1 != string::npos) return s.substr(loc1);
  return "0";
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a / gcd(a, b)) * b;
}
bool is_prime(long long int x) {
  if (x == 1) return false;
  if (x == 2 || x == 3) return true;
  for (long long int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
bool isp(string str) {
  int l = 0, h = str.length() - 1;
  while (h > l)
    if (str[l++] != str[h--]) return false;
  return true;
}
long long int power(long long int a, long long int b) {
  long long int res = 1ll;
  while (b > 0) {
    if (b % 2ll) res = (res * a) % 1000000007;
    ;
    a = (a * a) % 1000000007;
    ;
    b /= 2ll;
  }
  return res;
}
long long int interact(long long int i) {
  cout << i << "\n";
  cout.flush();
  long long int ret;
  cin >> ret;
  cout.flush();
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  long long int c = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) c++;
  }
  if (c) {
    cout << n - c << endl;
    return 0;
  }
  c = 1e10;
  for (long long int i = 0; i < n; i++) {
    long long int gc = a[i];
    for (long long int j = i + 1; j < n; j++) {
      gc = gcd(gc, a[j]);
      if (gc == 1) c = min(c, j - i);
    }
  }
  if (c == 1e10)
    cout << -1 << endl;
  else
    cout << c + n - 1 << endl;
  return 0;
}
