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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <math.h>
#include <numeric>
#include <complex>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <functional>
#include <assert.h>
using ll=long long;
#define int ll
#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define eb emplace_back
#define all(x) x.begin(),x.end()
#define SZ(x) ((int)x.size())
#define fi first
#define se second
const int INF = 1e9;
using namespace std;

template<class T> using vc=vector<T>;
template<class T> using vvc=vc<vc<T>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class T>
inline bool chmax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<class T>
inline bool chmin(T &a,T b){if(a>b){a=b;return true;}return false;}

signed main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout<<fixed<<setprecision(20);

    int N; cin >> N;
    int A[N+1];
    rep(i,N+1) cin >> A[i];
    int maxA[N+1];
    int pos=1;
    bool can=true;
    rep(i,N+1) {
        maxA[i]=min(pos,INF);
        if(pos<A[i]) {
            can=false;
            break;
        }
        pos-=A[i];
        pos*=2;
    }
    if(!can) {
        cout << -1 << endl;
        return 0;
    }
    int minT=0,maxT=0;
    int ans=0;
    per(i,N+1) {
        if(i==N){
            ans+=A[i];
            maxT=A[i];
        }
        else {
            ans+=min(maxT+A[i],maxA[i]);
            maxT+=A[i];
            chmin(maxT,maxA[i]);
        }
    }
    cout << ans << endl;

}