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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void) {
	int n[100001] = { 0 };
	int an[100001] = { 0 };
	int num = 0;

	for (int i = 0; i < 100001; i++) {
		n[i] = 2;
	}
	n[0] = 1;

	scanf("%d", &num);
	for (int i = 0; i <= num; i++) {
		int tmp = 0;
		scanf("%d", &tmp);
		an[i] = tmp;
	}

	int aa = 1;
	for (int i = 1; i <= num; i++) {
		aa *= 2;
	}

	int bb = 2;
	for (int i = 0; n[num] != aa; i++) {
		if (n[i] == an[i]) {
			break;
		}
		else {
			if (i == num) {
				for (int j = 0; j <= num; j++) {
					if (j == bb) {
						n[j] += 1;
					}
					else if (i > bb) {
						n[j] += 2;
					}
				}
				i = bb;
				bb++;
			}
		}
	}

	if ((n[num] == aa)||(n[num]+1 == aa)) {
		printf("-1");
	}
	else {
		int result = 0;
		for (int i = 0; i <= num; i++) {
			result += n[i];
		}
		printf("%d", result);
	}

	return 0;
}