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 JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

"use strict";
var input = require("fs").readFileSync("/dev/stdin", "utf8");
var cin = input.trim().split(/ |\n/), cid = 0;

function main() {
    let N = +cin.shift();
    let A = cin.map(x => +x);
    let node = Array(N + 1);
    let ans = 1;
    node[0] = 1;
    for (let d = 0; d < N; d++) {
        node[d + 1] = (node[d] - A[d]) * 2;
        if (node[d + 1] > 0) {
            ans += node[d + 1];
        } else {
            console.log(-1);
            return;
        }
    }
    //console.log(ans);
    if (node[N] < A[N]) {
        console.log(-1);
    } else if (node[N] == A[N]) {
        console.log(ans);
    } else {
        // node[d] が多すぎた場合。枝を逆に減らしていく必要がある。
        let over = Array(N + 1);
        let nodemax = Array(N + 1);
        nodemax[N] = A[N];
        for (let d = N; d > 0; d--) {
            //console.log({ d, A, node, nodemax, ans });
            if (nodemax[d] >= node[d]) { break; }
            over[d] = node[d] - nodemax[d]; // d層目で多すぎた葉
            ans = ans - over[d];
            nodemax[d - 1] = nodemax[d] + A[d - 1];
        }
        console.log(ans);
    }
}
main();