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>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

int n;
ll a[200005];
ll a2[200005];
ll sum[200005];
ll sum2[200005];

int main(void){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%lld",&a[i]);
        assert(a[i]>=0LL);
        a2[i]=-a[i]*2LL;
    }
    ll prev=a[n-1];
    stack<P> st;
    st.push(P(1,0));
    for(int i=n-2;i>=0;i--){
        sum[i]+=sum[i+1];
        ll len=0;
        while(prev<a[i]){
            while(st.size()>0){
                if(st.top().second==0LL){
                    len+=st.top().first;
                    st.pop();
                }else{
                    st.top().second--;
                    break;
                }
            }
            if(st.size()>0){
                sum[i]+=(len+st.top().first)*2LL;
            }else{
                sum[i]+=len*2LL;
            }
            //printf("go i %d %lld %lld %lld %d\n",i,prev,len,sum[i],st.size()==0?-1:st.top().first);
            prev*=4LL;
        }
        if(len>0){
            st.push(P(len,0));
        }
        ll cnt=0;
        ll po=a[i];
        while(prev>=po){
            cnt++;
            po*=4LL;
        }
        cnt--;
        if(cnt==0LL){
            st.top().first++;
        }else{
            st.top().second+=cnt;
            st.push(P(1,0));
        }
        //printf("push %lld\n",cnt);
        prev=a[i];
    }
    //printf(" go\n");
    for(int i=1;i<=n;i++){
        sum2[i]=1;
    }
    while(st.size())st.pop();
    prev=a2[0];
    st.push(P(1,0));

    for(int i=1;i<n;i++){
        sum2[i+1]+=sum2[i];
        ll len=0;
        while(prev>a2[i]){
            while(st.size()>0){
                if(st.top().second==0LL){
                    len+=st.top().first;
                    st.pop();
                }else{
                    st.top().second--;
                    break;
                }
            }
            //printf("%d %lld %lld\n",i,prev,len);
            if(st.size()>0){
                sum2[i]+=(len+st.top().first)*2LL;
            }else{
                sum2[i]+=len*2LL;
            }
            prev*=4LL;
        }
        if(len>0){
            st.push(P(len,0));
        }
        ll cnt=0;
        ll po=a2[i];
        while(prev<=po){
            cnt++;
            po*=4LL;
        }
        cnt--;
        if(cnt==0LL){
            st.top().first++;
        }else{
            st.top().second+=cnt;
            st.push(P(1,0));
        }
        //printf("push %lld\n",cnt);
        prev=a2[i];
    }

    ll ans=sum[0]+sum2[0];
    for(int i=0;i<=n;i++){
        //printf("i %d %lld %lld\n",i,sum[i],sum2[i]);
        if(sum[i]<0 || sum2[i]<0)continue;
        ans=min(ans,sum[i]+sum2[i]);
    }
    printf("%lld\n",ans);
    return 0;
}