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 mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
void solve();
void read_input() {}
void solve() {}
int gcd(int big, int small) {
  if (small > big) swap(big, small);
  int temp;
  while (small != 0) {
    if (small > big) swap(big, small);
    temp = big % small;
    big = small;
    small = temp;
  }
  return big;
}
const int MAXN = 2002;
int dp[MAXN][MAXN];
int a[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  bool hasone = false;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) hasone = true;
  }
  if (hasone) {
    int counts = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] != 1) {
        counts++;
      }
    }
    cout << counts << endl;
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    dp[i][i + 1] = gcd(a[i], a[i + 1]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 2; j < n; j++) {
      dp[i][j] = gcd(dp[i][j - 1], dp[j - 1][j]);
    }
  }
  int mindis = INT_MAX;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (dp[i][j] == 1) {
        mindis = min(mindis, j - i - 1);
      }
    }
  }
  if (mindis == INT_MAX)
    cout << -1 << endl;
  else
    cout << (mindis + n) << endl;
  return 0;
}
