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: Accepted)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i = 0;i < (int)(n);i++)
using ll = long long;
const ll MOD=1000000007;
const long long INF = 1LL << 60;
const double pi=acos(-1.0);

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; }

ll my_pow(long long x, long long n) {
    long long ret = 1;
    while (n > 0) {
        if (n & 1) ret = ret * x;  // n の最下位bitが 1 ならば x^(2^i) をかける
        x = x * x;
        n >>= 1;  // n を1bit 左にずらす
    }
    return ret;
}

int main()
{
  ll N; cin>>N;
  vector<ll> A(N+1); rep(i,N+1) cin>>A[i];
  
  //上から抑える
  vector<ll> msum(N+1,1);
  if(msum[0]-A[0]<0) {cout<<-1<<endl; return 0;}
  rep(i,N){
    msum[i+1]=min(INF,(msum[i]-A[i])*2);
    if(msum[i+1]-A[i+1]<0) {cout<<-1<<endl; return 0;}
  }
  
  //下から抑える
  ll sum=0,ans=0;
  for(int i=N;i>=0;i--){
    sum=min(sum+A[i],msum[i]);
    ans+=sum;
  }
  cout<<ans<<endl;
  return 0;
}