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;
typedef pair<int,int> iint;
const ll zero = 0;
const ll INF = 3000000000000000000; //10^18
const int inINF = 1000000000; //10^9
const ll MOD = 1000000007; //10^9+7
const ll MOD2 = 998244353;

int main(){
    ll N; cin >> N;
    vector<ll> A(N+1), B(N+1);
    ll sumA = 0;
    for (ll i = 0; i < N+1; i++) {
        cin >> A[i];
        sumA += A[i];
    }
    if(N == 0){
        if(A[0] == 1){
            printf("1\n");
            return 0;
        }
        else{
            printf("-1\n");
            return 0;
        }
    }

    if(A[0] != 0){
        printf("-1\n");
        return 0;
    }

    B[0] = 1;
    ll sumP = 1;

    for (ll i = 1; i < N+1; i++) {
        if(sumA - sumP >= B[i-1] - A[i-1]){
            B[i] = 2 * (B[i-1] - A[i-1]);
            sumP += B[i-1] - A[i-1];
        }
        else if (sumA == sumP){
            B[i] = B[i-1] - A[i-1];
        }
        else{
            B[i] = sumA - sumP + B[i-1] - A[i-1];
            sumP = sumA;
        }

        
        if(B[i] < A[i]){
            printf("-1\n");
            return 0;
        }
    }

    ll ans = 0;
    for (ll i = 0; i < N+1; i++) {
        ans += B[i];
        
    }
    printf("%lld\n", ans);

}