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;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int tree[200005];
long long int a[200005];
void build(long long int node, long long int start, long long int end) {
  if (start == end) {
    tree[node] = a[start];
  } else {
    int mid = (start + end) / 2;
    build(2 * node, start, mid);
    build(2 * node + 1, mid + 1, end);
    tree[node] = gcd(tree[2 * node], tree[2 * node + 1]);
  }
}
long long int query(long long int node, long long int start, long long int end,
                    long long int l, long long int r) {
  if (r < start or end < l) {
    return 0;
  }
  if (l <= start and end <= r) {
    return tree[node];
  }
  long long int mid = (start + end) / 2;
  long long int p1 = query(2 * node, start, mid, l, r);
  long long int p2 = query(2 * node + 1, mid + 1, end, l, r);
  return gcd(p1, p2);
}
int main() {
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, 1, n);
  long long int ans = 1000000007;
  long long int ones = 0;
  for (long long int i = 1; i <= n; i++) {
    if (a[i] == 1) {
      ones++;
    }
    for (long long int j = i + 1; j <= n; j++) {
      if (query(1, 1, n, i, j) == 1) {
        ans = min(ans, j - i);
      }
    }
  }
  if (ones == n) {
    cout << 0;
    return 0;
  }
  if (ans == 1000000007) {
    cout << -1;
    return 0;
  }
  cout << ans + n - 1 - ones;
}
