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;
vector<int> cnt;
int gcd(int x, int y) { return x == 0 ? y : gcd(y % x, x); }
const int maxn = 2005;
int dp[maxn][25];
void RMQ_init(int num) {
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < num; i++) dp[i][0] = cnt[i];
  for (int j = 1; (1 << j) <= num; j++) {
    for (int i = 0; i + (1 << j) - 1 < num; i++) {
      dp[i][j] = gcd(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
    }
  }
}
int RMQ(int L, int R) {
  int k = log(R - L + 1) / log(2);
  return gcd(dp[L][k], dp[R - (1 << k) + 1][k]);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    cnt.push_back(x);
  }
  int sz = cnt.size();
  RMQ_init(sz);
  if (RMQ(0, sz - 1) != 1) {
    printf("-1\n");
  } else {
    int cur = 0;
    for (auto it : cnt) {
      if (it == 1) cur++;
    }
    int minn = n * 10;
    for (int i = 0; i < sz; i++) {
      for (int j = i; j < sz; j++) {
        if (RMQ(i, j) == 1) minn = min(minn, j - i);
      }
    }
    if (cur == 0) cur = 1;
    printf("%d\n", minn + sz - cur);
  }
  return 0;
}
