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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define mod 998244353
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fll
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int maxn=1e5+10;
ll ned[maxn];
int n;
ll leavel[maxn],leaver[maxn],pw2[maxn];
int main()
{
	srand(time(0));
	pw2[0]=1;
	for(int i=1;i<=60;i++)
		pw2[i]=pw2[i-1]*2ll;
	for(int i=61;i<maxn;i++)pw2[i]=pw2[i-1];
//	cout<<pw2[maxn-1]<<endl;
	scanf("%d",&n);
//	n=1000;
	for(int i=0;i<=n;i++)
	{
	//	ned[i]=rand()%100;
		scanf("%lld",&ned[i]);
	}
//	ned[n]=rand()+1;
	leavel[n]=leaver[n]=ned[n];
	if(ned[n]>pw2[n])
	{
		printf("-1\n");
		return 0;
	}
	
	for(int lvl=n-1;lvl>=0;lvl--)
	{
		leavel[lvl]=(leavel[lvl+1]+1ll)/2ll;
		leaver[lvl]=leaver[lvl+1];
		
		leavel[lvl]+=ned[lvl];
		leaver[lvl]+=ned[lvl];
		
		leaver[lvl]=min(leaver[lvl],pw2[lvl]);
		if(leavel[lvl]>leaver[lvl])
		{
			printf("-1\n");
			return 0;
		}
	//	cout<<lvl<<" "<<leavel[lvl]<<" "<<leaver[lvl]<<endl;
	}
	ll all=leaver[0],nxt=leaver[0]-ned[0];
	for(int i=1;i<=n;i++)
	{
		nxt*=2ll;
		nxt=min(nxt,leaver[i]);
		all+=nxt;
		if(all<0)
		{
			exit(-1);
		}
		if(nxt<ned[i])
		{
			printf("-1\n");
			return 0;
		}
	//	cout<<i<<" "<<nxt<<" "<<nxt-ned[i]<<" "<<leavel[i]<<" "<<leaver[i]<<" "<<ned[i]<<" "<<pw2[i]<<endl;
	//	system("pause");
		nxt-=ned[i];
	}
//	cout<<nxt<<endl;
	all+=nxt;
	printf("%lld\n",all);
	return 0;
}