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)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
const ll mod = 1e9+7;
const int N = 2e5+5;
ll a[N];
ll b[N];
void solve(){
    int n;
    cin>>n;
    for(int i=0;i<=n;i++)cin>>a[i];
    b[n] = a[n];
    for(int i=n-1;i>=0;i--)b[i] = b[i+1]+a[i];
    ll ans = 0;
    ll nl = 0;
    ll l = 0;
    nl = 1;
    if(n==0){
        if(a[0]!=1){
            cout<<-1<<endl;
        }
        else{
            cout<<1<<endl;
        }
        return ;
    }
    if(a[0]==1){
        if(n>0){
            cout<<-1<<endl;
        }
        else{
            cout<<1<<endl;
        }
        return ;
    }
    for(int i=1;i<=n;i++){
        if(nl*2ll<a[i]){
            cout<<-1<<endl;
            return;
        }
        nl = nl*2;
        nl = min(nl,b[i]);
        ans+=nl;
        nl-=a[i];
    }
    cout<<ans+1<<endl;
}
int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t=1;
//    scanf("%d",&t);
    while(t--){
        solve();
    }
    return 0;
}