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 double EPS = 1e-9;
const double PI = 3.14159265359;
vector<long long> primes;
void GeneratingPrimes(int top) {
  vector<bool> p(top + 1, 1);
  p[0] = p[1] = 0;
  for (int i = 2; i <= sqrt(top); i++) {
    if (!p[i]) continue;
    if (p[i]) {
      for (int j = i * i; j <= top; j += i) {
        p[j] = 0;
      }
    }
  }
  for (int i = 0; i < top + 1; i++)
    if (p[i]) primes.push_back(i);
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void EXTgcd(long long a, long long b, int& x0, int& y0) {
  if (b == 0) {
    x0 = 1, y0 = 0;
    return;
  }
  EXTgcd(b, a % b, x0, y0);
  long long x1 = y0;
  long long y1 = x0 - (a / b) * y0;
  x0 = x1;
  y0 = y1;
}
long long POW(long long x, long long p) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret *= x;
    x *= x;
    p /= 2;
  }
  return ret;
}
double dist(int a, int b, int x, int y) {
  return sqrt(pow(a - x, 2) + pow(b - y, 2));
}
int xadd[9] = {1, -1, 0, 1, -1, 0, 1, -1, 0};
int yadd[9] = {1, -1, 0, -1, 0, 1, 0, 1, -1};
int n, ans = int(1e9);
vector<int> v(2001);
int main() {
  cin >> n;
  int O = 0;
  for (int i = 0; i < n; i++) cin >> v[i], O += (v[i] == 1);
  if (O) {
    cout << n - O;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    int l = v[i], al = 0, r = v[i], ar = 0;
    for (int j = i + 1; j < n; j++) {
      if (l == 1)
        break;
      else {
        l = gcd(v[j], l);
        al++;
      }
    }
    if (l == 1) {
      al += (n - 1);
    } else
      al = int(1e9);
    for (int j = i - 1; j >= 0; j--) {
      if (r == 1)
        break;
      else {
        r = gcd(v[j], r);
        ar++;
      }
    }
    if (r == 1) {
      ar += (n - 1);
    } else
      ar = int(1e9);
    ans = min(ans, min(al, ar));
  }
  if (ans == int(1e9))
    cout << -1;
  else
    cout << ans;
  return 0;
}
