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;
const long long INF = 1000000007;
const long long INFL = 1e15;
void printv(vector<int> a) {
  for (int i = 0; i < (int)a.size(); i++) {
    cout << a[i] << " ";
  }
}
long long fact_pow(long long n, long long k) {
  long long res = 0;
  while (n) {
    n /= k;
    res += n;
  }
  return res;
}
vector<int> allprime(long long n) {
  vector<char> prime(n + 1, true);
  prime[0] = prime[1] = false;
  for (int i = 2; i <= n; ++i)
    if (prime[i])
      if (i * 1ll * i <= n)
        for (int j = i * i; j <= n; j += i) prime[j] = false;
  vector<int> ans;
  for (int i = 2; i < n + 1; i++) {
    if (prime[i]) ans.push_back(i);
  }
  return ans;
}
long long binpow(long long a, long long n, long long m = 1e18 + 9) {
  if (n == 0) return 1;
  if (n % 2 == 1)
    return (binpow(a, n - 1) * a);
  else {
    long long b = binpow(a, n / 2);
    return (b * b);
  }
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
int n, mi = INF;
vector<int> a;
void up(int j) {
  long long d = a[j];
  int x = 0;
  for (int i = j - 1; i > -1; i--) {
    d = gcd(a[i], d);
    x++;
    if (d == 1) break;
  }
  if (d == 1) mi = min(x, mi);
}
void down(int j) {
  long long d = a[j];
  int x = 0;
  for (int i = j + 1; i < n; i++) {
    d = gcd(d, a[i]);
    x++;
    if (d == 1) break;
  }
  if (d == 1) mi = min(x, mi);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  a.resize(n);
  int c1 = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) c1++;
  }
  if (c1) {
    cout << n - c1;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    up(i);
    down(i);
  }
  if (mi == INF)
    cout << -1;
  else
    cout << mi + n - 1;
  return 0;
}
