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 <float.h>
#include <inttypes.h>
#include <limits.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif

#define getchar getchar_unlocked
#define putchar putchar_unlocked
int32_t nextint(void){ char c=getchar_unlocked(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar_unlocked(); int s=0; if(c=='-'){s=1;c=getchar_unlocked();} uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar_unlocked(); } return s?-x:x; }

int a[100001];
int b[100001];
int main(void){
	int n=nextint();
	a[0]=nextint();
	if(n==0){
		if(a[0]==1){
			puts("1");
		}else{
			puts("-1");
		}
	}else{
		if(a[0]==0){
			b[0]=1;
			for(int i=1; i<=n; i++){
				a[i]=nextint();
				if(a[i]>b[i-1]*2){
					puts("-1");
					return 0;
				}
				b[i]=b[i-1]*2-a[i];
				if(b[i]>INT_MAX/2) b[i]=INT_MAX/2;
			}
			b[n]=0;
			for(int i=n; i>0; i--){
				if(b[i-1]>=a[i]+b[i]){
					b[i-1]=a[i]+b[i];
				}
			}
			long ans=0;
			for(int i=0; i<=n; i++){
				ans+=a[i];
				ans+=b[i];
			}
			printf("%ld\n", ans);
		}else{
			puts("-1");
		}
	}
}