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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define rep(i,n)for(lld i=0;i<(lld)(n);i++)
typedef long long lld;
typedef long double Lf;
#define Ma(x,y) (x>y?x:y)
#define mi(x,y) (x<y?x:y)
#define N 100201

int main() {
	lld n,a[N],i,next=0,ans=0;
	scanf("%lld",&n);
	rep(i,n+1)scanf("%lld",&a[i]);
	if (n == 0 && a[0] != 1) { printf("-1"); return 0; }
	if (n == 0 && a[0] == 1) { printf("1"); return 0; }
	if (a[0] != 0) { printf("-1"); return 0; }
	if (n == 1 && a[1] == 1) { printf("2"); return 0; }
	if (n == 1 && a[1] == 2) { printf("3"); return 0; }
	if (n == 1 && a[1] > 2) { printf("-1"); return 0; }
	lld b[N] = { 0 };
	b[0] = 1;
	rep(i, n) {
		b[i + 1] = b[i] * 2 - a[i + 1];
		if(b[i+1]<0){ printf("-1"); return 0; }
		if(i!=n-1&&b[i+1]==0) { printf("-1"); return 0; }
	}
	for (i=n;i>=1;i--){
		ans += next+a[i];
		next = mi(next+a[i], b[i-1]);
		if (next==0&&a[i-1]==0&&i>1) { printf("-1"); return 0; }
	}
	printf("%lld", ans+1);
}