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)


// **** WHEN THE GOING GETS TOUGH, THE TOUGH GETS GOING. ****

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define ll long long int
#define ull unsigned long long int
#define ld long double
#define ordered_set tree < ll ,  null_type ,  less<ll> ,  rb_tree_tag ,  tree_order_statistics_node_update >
//added two more functions to set
//(1)*(set.find_by_order(k))[kth element in the sorted set] 
//(2)set.order_of_key(k)[count of elements strictly less than k]

typedef vector< int > vi;
typedef vector< ll > lvi;
typedef vector< vi > vvi;
typedef vector< lvi > lvvi;
typedef pair< int,int > ii;
typedef pair< ll,ll > lii;
typedef vector<ii> vii;
typedef vector<lii> lvii;

#define min_prq(t) priority_queue<t,vector<t>,greater<t>>
#define max_prq(t) priority_queue<t>
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(auto i = (c).begin(); i != (c).end(); i++)//remember i is an iterator
#define present(c,x) ((c).find(x) != (c).end())//for sets,maps,multimaps
#define cpresent(c,x) (find(all(c),x) != (c).end())//for vectors
#define tc(t) int (t); cin>>(t);while((t)--)
#define ff first
#define ss second
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cerr << *it << " is " << a << endl;
	err(++it, args...);
}

//-----------------------------------------------------------------------------------------------------------------------------------------------------------------
#define mod 1000000007
//-----------------------------------------------------------------------------------------------------------------------------------------------------------------

ll mod_sum() { return 0LL; }
template<typename T, typename... Args>
T mod_sum(T a, Args... args) { return ((a + mod_sum(args...))%mod + mod)%mod; }

ll mod_product() { return 1LL; }
template<typename T, typename... Args>
T mod_product(T a, Args... args) { return (a*mod_product(args...))%mod; }
//-----------------------------------------------------------------------------------------------------------------------------------------------------------------


signed main()
{
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);	
	ll n;
	cin>>n;
	lvii nodes(n+2,{1,1});
	lvi a(n+1);
	bool flag=true;
	for (int i = 0; i < n+1; ++i)
	{
		cin>>a[i];
		//checking if free nodes at level i is greater than the  max nodes at that level
		if(a[i]>nodes[i].ss) flag=false;
		//



		if(a[i]>=nodes[i].ff) nodes[i+1].ff=0;
		else nodes[i+1].ff=(nodes[i].ff-a[i]);



		nodes[i+1].ss=(nodes[i].ss-a[i])*2;
		// error(nodes[i].ff,nodes[i].ss);
	}
	if(a[n]<nodes[n].ff) flag=false;
	if(a[n]==0) flag=false;
	if(!flag)
	{
		cout<<"-1"<<endl;
		return 0;
	}
	ll ans=a[n];
	ll ct=a[n];
	for(int i=n-1;i>=0;i--)
	{
		// if(((ct+1)/2 + a[i])>nodes[i].ss) flag=false;
		ct+=a[i];
		if(ct>nodes[i].ss) ct=nodes[i].ss;
		if(ct<nodes[i].ff) flag=false;
		ans+=ct;
		// error(ct);
	}
	if(!flag)
	{
		cout<<"-1"<<endl;
		return 0;
	}
	cout<<ans<<endl;
	return 0;
}