Problem:
Score : 800 points

Problem Statement

There are N positive integers A_1, A_2, ..., A_N.
Takahashi can perform the following operation on these integers any number of times:

Choose 1 \leq i \leq N and multiply the value of A_i by -2.

Notice that he multiplies it by minus two.

He would like to make A_1 \leq A_2 \leq ... \leq A_N holds.
Find the minimum number of operations required. If it is impossible, print -1.

Constraints

1 \leq N \leq 200000

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the answer.

Sample Input 1

4
3 1 4 1

Sample Output 1

3

One possible solution is:

Choose i=4 and multiply the value of A_4 by -2. A_1, A_2, A_3, A_4 are now 3, 1, 4, -2.

Choose i=1 and multiply the value of A_1 by -2. A_1, A_2, A_3, A_4 are now -6, 1, 4, -2.

Choose i=4 and multiply the value of A_4 by -2. A_1, A_2, A_3, A_4 are now -6, 1, 4, 4.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

0

A_1 \leq A_2 \leq ... \leq A_N holds before any operation is performed.

Sample Input 3

8
657312726 129662684 181537270 324043958 468214806 916875077 825989291 319670097

Sample Output 3

7

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

using int64 = long long;
const int64 INF = 1LL << 60;


vector< int64 > solve(vector< int64 > A) {
  int N = (int) A.size();
  stack< int > st;
  vector< int64 > result(N);
  int64 ret = 0;

  st.emplace(N - 1);
  for(int i = (int) A.size() - 2; i >= 0; i--) {

    while(A[i] > A[i + 1]) {
      A[i + 1] *= 4;
      int r = st.top();
      st.pop();
      if(i + 1 != r) A[r] *= 4;
      ret += (r - i) * 2;
      if(r + 1 == N || A[r] * 4 <= A[r + 1]) st.emplace(r);
    }
    if(A[i] * 4 <= A[i + 1]) st.emplace(i);
    result[i] = ret;
  }
  return result;
}


int main() {
  int N;
  cin >> N;
  vector< int64 > A(N);
  for(int i = 0; i < N; i++) {
    cin >> A[i];
  }
  auto suf = solve(A);
  for(int i = 0; i < N; i++) {
    A[i] *= 2;
  }
  reverse(begin(A), end(A));
  auto pre = solve(A);
  reverse(begin(pre), end(pre));
  int64 ret = min(pre[N - 1] + N, suf[0]);
  for(int i = 1; i < N; i++) {
    ret = min(ret, pre[i - 1] + suf[i] + i);
  }
  cout << ret << endl;
}