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: Wrong Answer)

#include <bits/stdc++.h>
 
using namespace std;
 
const int N = 1000 * 1000 + 17;
const int MOD = 1000 * 1000 * 1000 + 7;

int n;
long long a[N], ldp[N], rdp[N];

bool read() {
    if (!(cin >> n))
        return false;

    for (int i = 1; i <= n; ++i)
        cin >> a[i];

    return true;
}

void solve() {
    for (int i = 1; i <= n; ++i) {
        auto k = 0ll;
        for (auto x = a[i - 1]; i != 1 && x < a[i]; x *= 4, ++++k) { }
        ldp[i] = 1 + ldp[i - 1] + k * (i - 1);
    }

    for (int i = n; i >= 1; --i) {
        auto k = 0ll;
        for (auto x = a[i + 1]; i != n && x < a[i]; x *= 4, ++++k) { }
        rdp[i] = rdp[i + 1] + k * (n - i);
    }

    auto ans = 1ll * MOD * MOD;

    for (int i = 0; i <= n; ++i)
        ans = min(ans, ldp[i] + rdp[i + 1]);

    cout << ans << endl;
}
 
int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    while (read())
        solve();

    return 0;
}