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)

////////////////////////////////////////////
///                                      ///
///        Template ver. 1.2. rapel      ///
/// Fear is Temporary, Regret is Forever ///
///          Must Try and Get AC         ///
///                                      ///
////////////////////////////////////////////
#include <stdio.h>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <utility>

#include <math.h>
#include <complex>

#include <assert.h>
#include <time.h>
#include <chrono>
#include <random>

#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <list>
#include <bitset>

#define F first
#define S second
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define endl '\n'
using namespace std;

typedef long long ll;
typedef unsigned long long ull;

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void desperate_optimization(int precision){
  	ios_base::sync_with_stdio(false);
  	cin.tie(0);
  	cout.tie(0);
  	cout.setf(ios::fixed);
  	cout.setf(ios::showpoint);
  	cout.precision(precision);
}

const int N = 1e5;
ll a[N + 5];
ll ssum[N + 5];

void hora(int tc) {
	int n;
	cin>>n;
	for(int i = 0;i <= n;i++) cin>>a[i];
	if(n == 0) {
		if(a[0] == 1) cout<<1<<endl;
		else cout<<-1<<endl;
		return ;
	}
	for(int i = n;i >= 0;i--) ssum[i] = ssum[i + 1] + a[i];
	ll cur = 1;
	bool bisa = true;
	for(int i = 0;i <= n;i++) {
		cur -= a[i];
		if(cur < 0) bisa = false;
		cur *= 2;
	}
	if(bisa == false) {
		cout<<-1<<endl;
		return ;
	}
	ll tot = 0;
	cur = 1;
	for(int i = 0;i < n;i++) {
		cur -= a[i];
		if(cur * 2 <= ssum[i + 1]) {
			tot += cur * 2;
			cur *= 2;
		} else {
			tot += ssum[i + 1];
			cur = max(cur, ssum[i + 1]);
		}
	}
	cout<<tot + 1<<endl;
}

int main(){
	desperate_optimization(10);
	int ntc = 1;
	//cin>>ntc;
	for(int tc = 1;tc <= ntc;tc++) hora(tc);
	return 0;
}