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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define ll long long

int main() {
  int n=0;
  cin>>n;
  vector<int> vec(n+1);
  rep(i,n+1){
    cin>>vec.at(i);
  }
  vector<pair<int,int>> howLayer(n+1);
  ll count=n+1;
  int nowLayer=0;
  int countroop=0;
  
  rep(i,n+1){
    howLayer.at(i).first=1;
  }
  
  for(int j=1;j<n;j++){
    while(vec.at(j)!=0){
      count+=j-nowLayer;
      //cout<<nowLayer;
      vec.at(j)--;
      howLayer.at(nowLayer).second++;
      for(int i=nowLayer+1;i<j+1;i++){
        howLayer.at(i+1).first++;
      }
      if(howLayer.at(nowLayer).first==howLayer.at(nowLayer).second){
        nowLayer++;
      }
    }
  }
  
  while(vec.at(n)!=1){
    count+=n-nowLayer;
    //cout<<nowLayer;
    vec.at(n)--;
    howLayer.at(nowLayer).second++;
    for(int i=nowLayer+1;i<n+1;i++){
      howLayer.at(i).first++;
    }
    if(howLayer.at(nowLayer).first==howLayer.at(nowLayer).second){
      nowLayer++;
    }
  }
  cout<<count;
}