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 comment(linker, "/STACK:1710886400")
using namespace std;
const int MOD = 2000000033;
const int INF = 1000000001;
const long double EPS = 1e-15;
const int HASH_POW = 29;
const long double PI = acos(-1.0);
mt19937_64 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long rndNext(long long l, long long r) {
  uniform_int_distribution<long long> foo(l, r);
  return foo(rnd);
}
double workTime() { return double(clock()) / CLOCKS_PER_SEC; }
void my_return(int code) { exit(code); }
int gcd(int a, int b) {
  if (!b) {
    return a;
  }
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, a[2010];
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  int cnt1 = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) {
      ++cnt1;
    }
  }
  if (cnt1 > 0) {
    printf("%d\n", n - cnt1);
    my_return(0);
  }
  int ans = INF;
  for (int ll = 1; ll <= n; ++ll) {
    int val = a[ll], rr = ll;
    do {
      val = gcd(val, a[rr]);
      ++rr;
    } while (val != 1 && rr <= n);
    if (val == 1) {
      ans = min(ans, rr - ll + n - 2);
    }
  }
  if (ans == INF) {
    printf("-1\n");
  } else {
    printf("%d\n", ans);
  }
  my_return(0);
}
