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)

#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 1000000000
#define INF 9223372036854775807
#define MOD 998244353
#define mod 1000000007
#define INT_MAX_ 2147483647
#define EPS (1e-10)
#define REP(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
#define rep(i,n)for (ll i = 0; i < (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
#define me memset
#define bit(n,k) ((n>>k)&1)
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T> >;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp1(pair<ll,Pll> a,pair<ll,Pll> b){
        if(a.fi!=b.fi)
        return a.fi<b.fi;
        else
        return a.se.fi>b.se.fi;
}
//----------------------------------------------------------------------
ll cal[100010];
ll can[100010];
//----------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------  
    //ll begin_t=clock();
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //-------------------------------
    ll n;cin>>n;
    V<ll>a(n+1);
    ll sum=0;
    for(ll i=0;i<=n;i++){
        cin>>a[i];
        sum+=a[i];
    }
    can[0]=1;
    for(ll i=0;i<=n;i++){
        if(i)can[i]=mmin(2*can[i-1],100000000);
        if(can[i]<a[i]){
            cout<<-1<<endl;
            return 0;
        }
        sum-=a[i];
        if(can[i]==a[i]&&i<n){
            cout<<-1<<endl;
            return 0;
        }
        if(i){
            if(a[i]>can[i-1]*2){
                cout<<-1<<endl;
                return 0;
            }
        }
    }
    ll ans=a[n];
    ll pre=a[n];
    cal[n]=a[n];
    for(ll i=n-1;i>=0;i--){
        ll now = pre + a[i]; 
        chmin(now,can[i]);
        if(i){
            chmin(now,(can[i-1]-a[i-1])*2);
        }
        pre=now;
        cal[i]=now;
        if(cal[i]<0){
            cout<<-1<<endl;
            return 0;
        }
    }
    ans=1;
    pre=1;
    for(ll i=1;i<=n;i++){
        if(pre<a[i-1]){
            cout<<-1<<endl;
            return 0;
        }
        ll maxs=(pre-a[i-1])*2;
        chmin(maxs,cal[i]);
        ans+=maxs;
        pre=maxs;
    }
    cout<<ans<<endl;
    //-------------------------------  
    //fclose(stdin);
    //fclose(stdout);
    //ll end_t=clock();cout<<"time="<<end_t-begin_t<<"ms"<<endl;
    //-------------------------------
    return 0;
}
//----------------------------------------------------------------------