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>
#pragma GCC optimize("O3")
using namespace std;
int in() {
  int x;
  cin >> x;
  return x;
}
long long lin() {
  long long x;
  cin >> x;
  return x;
}
string stin() {
  string s;
  cin >> s;
  return s;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
vector<int> iota(int n) {
  vector<int> a(n);
  iota(begin(a), end(a), 0);
  return a;
}
void print() { putchar(' '); }
void print(bool a) { cout << a; }
void print(int a) { cout << a; }
void print(long long a) { cout << a; }
void print(char a) { cout << a; }
void print(string& a) { cout << a; }
void print(double a) { cout << a; }
template <class T>
void print(const vector<T>&);
template <class T, size_t size>
void print(const array<T, size>&);
template <class T, class L>
void print(const pair<T, L>& p);
template <class T, size_t size>
void print(const T (&)[size]);
template <class T>
void print(const vector<T>& a) {
  if (a.empty()) return;
  print(a[0]);
  for (auto i = a.begin(); ++i != a.end();) {
    cout << " ";
    print(*i);
  }
  cout << endl;
}
template <class T>
void print(const deque<T>& a) {
  if (a.empty()) return;
  print(a[0]);
  for (auto i = a.begin(); ++i != a.end();) {
    cout << " ";
    print(*i);
  }
}
template <class T, size_t size>
void print(const array<T, size>& a) {
  print(a[0]);
  for (auto i = a.begin(); ++i != a.end();) {
    cout << " ";
    print(*i);
  }
}
template <class T, class L>
void print(const pair<T, L>& p) {
  cout << '(';
  print(p.first);
  cout << ",";
  print(p.second);
  cout << ')';
}
template <class T, size_t size>
void print(const T (&a)[size]) {
  print(a[0]);
  for (auto i = a; ++i != end(a);) {
    cout << " ";
    print(*i);
  }
}
template <class T>
void print(const T& a) {
  cout << a;
}
int out() {
  putchar('\n');
  return 0;
}
template <class T>
int out(const T& t) {
  print(t);
  putchar('\n');
  return 0;
}
template <class Head, class... Tail>
int out(const Head& head, const Tail&... tail) {
  print(head);
  putchar(' ');
  out(tail...);
  return 0;
}
long long gcd(long long a, long long b) {
  while (b) {
    long long c = b;
    b = a % b;
    a = c;
  }
  return a;
}
long long lcm(long long a, long long b) {
  if (!a || !b) return 0;
  return a * b / gcd(a, b);
}
template <class... T>
void err(const T&...) {}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int n = in();
  vector<int> a;
  for (long long i = 0; i < n; i++) a.push_back(in());
  if (n == 1) {
    if (a[0] == 1)
      cout << 0 << '\n';
    else
      cout << -1 << '\n';
    return 0;
  }
  int mi = INT_MAX;
  for (long long i = 0; i < n; i++) {
    int g = a[i];
    for (long long j = 1; j <= n; ++j) {
      if (i + j == n) break;
      g = gcd(g, a[i + j]);
      if (g == 1) {
        chmin(mi, (int)j + 1);
        break;
      }
    }
  }
  int cnt = count(begin(a), end(a), 1);
  if (cnt >= 1) {
    cout << n - cnt << '\n';
    return 0;
  }
  cout << (mi == INT_MAX ? -1 : mi + n - 2) << '\n';
}
