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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define int long long
using namespace std;

int a[100010];
int ll[100010], rr[100010];

signed main () {
	int n; scanf("%d", &n);
	for (int i = 0; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	if (n == 0) {
		if (a[0] == 1) printf("1\n");
		else printf("-1\n");
		return 0;
	}
	if (a[0] != 0) {
		printf("-1\n");
		return 0;
	}
	int l = 1, r = 1;
	ll[0] = l, rr[0] = r;
	for (int i = 1; i <= n; i++) {
		if (r * 2 < a[i]) {
			printf("-1\n");
			return 0;
		}
		r = r * 2 - a[i];
		l = max(0ll, l - a[i]);
		ll[i] = l, rr[i] = r;
	}
	ll[n] = max(ll[n], 0ll), rr[n] = min(0ll, rr[n]);
	if (ll[n] > rr[n]) {
		printf("-1\n");
		return 0;
	}
	long long ans = rr[n] + a[n];
	for (int i = n - 1; i >= 0; i--) {
		ll[i] = max(ll[i], (ll[i + 1] + a[i + 1] + 1) / 2);
		rr[i] = min(rr[i], rr[i + 1] + a[i + 1]);
		if (ll[i] > rr[i]) {
			printf("-1\n");
			return 0;
		}
		ans += rr[i] + a[i];
	}
	printf("%lld\n", ans);
	return 0;
}