Problem:
Score : 600 points

Problem Statement

Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.

Notes

A binary tree is a rooted tree such that each vertex has at most two direct children.

A leaf in a binary tree is a vertex with zero children.

The depth of a vertex v in a binary tree is the distance from the tree's root to v. (The root has the depth of 0.)

The depth of a binary tree is the maximum depth of a vertex in the tree.

Constraints

0 \leq N \leq 10^5

0 \leq A_i \leq 10^{8} (0 \leq i \leq N)

A_N \geq 1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_0 A_1 A_2 \cdots A_N

Output

Print the answer as an integer.

Sample Input 1

3
0 1 1 2

Sample Output 1

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print 7.

Sample Input 2

4
0 0 1 0 2

Sample Output 2

10

Sample Input 3

2
0 3 1

Sample Output 3

-1

Sample Input 4

1
1 1

Sample Output 4

-1

Sample Input 5

10
0 0 1 1 2 3 5 8 13 21 34

Sample Output 5

264

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
using Graph=vector<vector<int>>;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
typedef long long ll;
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

int main(){
    ll N;cin>>N;
    vector<ll>A(N+1);
    rep(i,N+1){
        cin>>A.at(i);
    }
    vector<ll>C(N+1);
    C.at(0)=1;
    rep(i,N){
        C.at(i+1)=2*C.at(i);
    }
    vector<ll>B(N+1);
    B.at(N)=A.at(N);
    rep(i,N){
        B.at(N-1-i)=B.at(N-i)+A.at(N-1-i);
    }
    ll sum=0;
    bool flag=true;
    rep(i,N+1){
        B.at(i)=min(B.at(i),C.at(i));
    }
    rep(i,N+1){
        B.at(i)=B.at(i)-A.at(i);
        if(B.at(i)<0){
            flag=false;
            break;
        }
        if(i!=N){
            B.at(i+1)=min(B.at(i)*2,B.at(i+1));
        }
    }
    rep(i,N){
        if(B.at(i)*2<B.at(i+1)){
            flag=false;
        }
        if(B.at(i)*2<B.at(i+1)+A.at(i+1)){
            flag=false;
        }
    }
    rep(i,N+1){
        sum=sum+B.at(i)+A.at(i);
    }
    if(flag){
        cout<<sum<<endl;
    }
    else{
        cout<<-1<<endl;
    }
}