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)

#define _USE_MATH_DEFINES
#define _SILENCE_CXX17_C_HEADER_DEPRECATION_WARNING
#define _SILENCE_ALL_CXX17_DEPRECATION_WARNINGS
#define _CRT_SECURE_NO_WARNINGS
//#include <bits/stdc++.h>

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#include <cassert>
#include <fstream>
#include <array>
#include <string>
#include <cstring>
#include <functional>
#include <ctime>
#include <iostream>
#include <unordered_set>
#include <complex>

#include <nmmintrin.h>
#include <mmintrin.h>
#include <emmintrin.h>


using namespace std;

#define dprint(Exp,...) if(Exp){fprintf(stderr, __VA_ARGS__);}
#define printe(...) fprintf(stderr, __VA_ARGS__);
//#define printe(...) void();
#define printE(...) fprintf(stderr, __VA_ARGS__);
//#define printE(...) void();


#define PrtExp(_Exp)  cerr<< #_Exp <<" = "<< (_Exp)
#define PrtExpN(_Exp)  cerr<< #_Exp <<" = "<< (_Exp) <<"\n"

#define SINT(n) scanf("%d",&n)
#define SINT2(n,m) scanf("%d %d",&n,&m)
#define SINT3(n,m,o) scanf("%d %d %d",&n,&m,&o)
#define SINT4(n,m,o,P) scanf("%d %d %d %d",&n,&m,&o,&P)
#define SINT5(n,m,o,P,q) scanf("%d %d %d %d %d",&n,&m,&o,&P,&q)
#define SLL(n) scanf("%lld",&n)
#define SLL2(n,m) scanf("%lld %lld",&n,&m)
#define SLL3(n,m,o) scanf("%lld %lld %lld",&n,&m,&o)
#define SST(s) scanf("%s",s)
#define SCH(c) scanf("%c",&c)

#define GC() getchar()

#define PINT(n) printf("%d",(int)(n))
#define PINT2(n,m) printf("%d %d",(int)(n),(int)(m))
#define PINT3(n,m,l) printf("%d %d %d",(int)(n),(int)(m),(int)(l))
#define PLL(n) printf("%lld",(long long)(n))
#define PST(s) printf("%s",(s))
#define PCH(s) printf("%c",(s))

#define PINTN(n) printf("%d\n",(int)(n))
#define PINT2N(n,m) printf("%d %d\n",(int)(n),(int)(m))
#define PINT3N(n,m,l) printf("%d %d %d\n",(int)(n),(int)(m),(int)(l))
#define PLLN(n) printf("%lld\n",(long long)(n))
#define PSTN(s) printf("%s\n",(s))
#define PCHN(s) printf("%c\n",(s))

#define PSP() printf(" ")
#define PN() printf("\n")

#define PC(c) putchar(c)
#define CSP (' ')
#define SN ("\n")

#define rep(i,a) for(int i=0;i<a;i++)
#define reP(i,a) for(int i=0;i<=a;i++)
#define Rep(i,a) for(int i=a-1;i>=0;i--)
#define ReP(i,a) for(int i=a;i>=0;i--)

#define rEp(i,a) for(i=0;i<a;i++)
#define rEP(i,a) for(i=0;i<=a;i++)
#define REp(i,a) for(i=a-1;i>=0;i--)
#define REP(i,a) for(i=a;i>=0;i--)

#define repft(i,a,b) for(int i=a;i<b;i++)
#define repfT(i,a,b) for(int i=a;i<=b;i++)
#define Repft(i,a,b) for(int i=a-1;i>=b;i--)
#define RepfT(i,a,b) for(int i=a;i>=b;i--)

#define foreach(a,it) for(auto it = a.begin(); it != a.end(); ++it)

#define FILL(a,v) fill(begin(a),end(a), v)
#define FILL0(a) memset(a,0,sizeof(a))
#define FILL1(a) memset(a,-1,sizeof(a))

typedef long long ll;

typedef unsigned long long ull;
typedef pair<int, int> Pi;
typedef pair<ll, ll>   Pll;


const int INF = 1'010'000'000; // 0x3C33'6080
const ll INFLL = 0x1f1f1f1f1f1f1f1fLL;//2,242,545,357,980,376,863


template <class A, class B> inline ostream& operator<<(ostream& st, const pair<A, B>& P) { return st << "(" << P.first << "," << P.second << ")"; };
template <class A, class B> inline pair<A, B> operator+(const pair<A, B>& P, const pair<A, B>& Q) { return pair<A, B>(P.first + Q.first, P.second + Q.second); };
template <class A, class B> inline pair<A, B> operator-(const pair<A, B>& P, const pair<A, B>& Q) { return pair<A, B>(P.first - Q.first, P.second - Q.second); };


inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }


int a[100005];

int main() {
    ll n;
    cin >> n;

    ll t = 0;
    reP(i, n) {
        cin >> a[i];
        t += a[i];
    }
    ll ret = 0;

    ll x = 1, d = 0;
    reP(i, n) {
        //printf("%d : z%d  x%lld  t%lld\n", i, a[i], x, t);

        if (a[i] > x) {
            cout << -1 << endl;
            return 0;
        }
        x -= a[i];
        t -= a[i];
        
        ret += a[i];
        ret += x;
        x = min(x * 2, t);
    }

    cout << ret << endl;



    return 0;
}