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;
#define rep(i,n) for(int i=0;i<(n);++i)
#define Rep(i,n) for(int i=1;i<=(n);++i)
using LL = long long;
using ULL = unsigned long long;

const int INF = 0x3f3f3f3f;
const double pi = acos(-1.0);

LL up[100123]{1}, a[100123], b[100123];

int main()
{
    Rep(i, 59) up[i] = up[i-1] * 2;
    int n;
    cin >> n;
    rep(i, n+1) scanf("%lld", a+i);
    b[n] = a[n];
    for(int i = n; i > 60; --i) b[i-1] = a[i-1] + (b[i] + 1 >> 1);
    for(int i = 59; i >= 0; --i) {
        b[i] = a[i] + (b[i+1] + 1 >> 1);
        if(b[i] > up[i]) {
            puts("-1");
            return 0;
        }
    }
    int p = n;
    LL cur = 1 - a[0];
    Rep(i, n) {
        up[i] = cur * 2;
        cur = up[i] - a[i];
        if(up[i] > 1e18) {
            p = i;
            break;
        }
    }
    for(int i = n; i > p; --i) b[i-1] = a[i-1] + b[i];
    for(int i = p-1; i >= 0; --i) {
        b[i] = min(a[i] + b[i+1], up[i]);
    }
    cout << accumulate(b, b+n+1, 0ll) << endl;
    return 0;
}