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;

#include <vector>
#include <map>
#include <set>
#include <queue>
#include <math.h>
#include <iostream>
#include <stdio.h>

typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pairll;

#define sll(x) scanf("%lld",&x);
#define READ freopen("in.txt","r",stdin)
#define WRITE freopen("out.txt","w",stdout)

#define ff first
#define ss second

inline ll readInt()
{
    ll in; scanf("%lld",&in); return in;
}

inline ll ADD(ll a, ll b, ll mod)
{
    return (a + b + mod + mod)%mod;
}

inline ll SUBS(ll a, ll b, ll mod)
{
    return (a-b+mod + mod) %mod;
}

inline ll MUL(ll a, ll b, ll mod)
{
    return (a*b + mod)%mod;
}

ll bigMod(ll base, ll power, ll mod)
{
    if(!power) return 1;
    if(!base) return 0;

    ll temp = bigMod(base, power/2, mod);

    if(power%2 == 0)
    {
        return (temp*temp)%mod;
    }
    else return (((temp*temp)%mod)*base)%mod;
}

inline ll modInverse(ll a, ll modp)
{
    return bigMod(a, modp -2, modp);
}

inline ll modInverseExtended(ll a, ll m) 
{ 
    ll m0 = m; 
    ll y = 0, x = 1; 
  
    if (m == 1) 
      return 0; 
  
    while (a > 1) 
    { 
        // q is quotient 
        ll q = a / m; 
        ll t = m; 
  
        // m is remainder now, process same as 
        // Euclid's algo 
        m = a % m, a = t; 
        t = y; 
  
        // Update y and x 
        y = x - q * y; 
        x = t; 
    } 
  
    // Make x positive 
    if (x < 0) 
       x += m0; 
  
    return x; 
} 
  

void autoinput()
{
    #ifndef ONLINE_JUDGE
    READ; 
    //WRITE;

    #endif
}

int main()
{
    autoinput();

    ll N;

    sll(N);

    vector<ll> leaves(N+1), freedom(N+1);

    freedom[0] = 1;

    for(ll i = 0; i <= N; i++)
    {
        sll(leaves[i]);
    }

    for(ll i = 0; i <= N; i++)
    {
        if(!i)
        {   
            freedom[i] -= leaves[i];
        }
        else
        {
            if(freedom[i-1] < INT64_MAX/10) freedom[i] = 2*freedom[i-1] - leaves[i];
            else freedom[i] = freedom[i-1];
        }

        if(freedom[i] < 0)
        {
            puts("-1"); return 0;
        }

    }

    vll tree(N+1);


    tree[N] = leaves[N];

    ll sum = tree[N];
    
    for(ll i = N-1; i >= 0; i--)
    {
        tree[i] = min(freedom[i],tree[i+1]);
        tree[i] += leaves[i];

        sum += tree[i];

    }

    printf("%lld\n",sum);

}