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; 

typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef vector<PII> VPII;
typedef long long LL;
typedef vector<LL> VL;
typedef vector<VL> VVL;
typedef pair<LL, LL> PLL;
typedef vector<PLL> VPLL;
typedef priority_queue<int> PQ_DESC;
typedef priority_queue<int, vector<int>, greater<int>> PQ_ASC;
typedef priority_queue<PII> PQ_DESC_PII;
typedef priority_queue<PII, vector<PII>, greater<PII>> PQ_ASC_PII;

#define ALL(c) (c).begin(),(c).end()
#define PB push_back
#define MP make_pair
#define SORT_ASC(c) sort(ALL(c))
//#define SORT_DESC(c) sort(ALL(c), greater<typeof(*((c).begin()))>())
#define SORT_DESC(c) sort((c).rbegin(),(c).rend())
#define REV(c) reverse((c).begin(), (c).end())
#define SIZE(a) int((a).size())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define ROF(i,a,b) for(int i=(b-1);i>=(a);--i)
#define REP(i,n) FOR(i,0,n)
#define PER(i,n) ROF(i,0,n)

const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int LARGE_INT = 1e9+100;
const int INF = 2e9+100;
const LL INF_LL = (LL)INF*(LL)INF;
const int MOD = 1e9+7;

//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

LL modpow(LL a, LL n) {
    LL res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}

void Main()
{
    LL n;cin>>n;
    VL a(n+1);
    REP(i,n+1){
        cin>>a[i];
    }
    if(n==0){
        if(a[0] == 1){
            cout<<1<<endl;
            return;
        }
        else{
            cout<<-1<<endl;
            return;
        }
        return;
    }

    VL c(n+1);
    c[n]=a[n];
    PER(i,n){
        c[i] = a[i]+c[i+1];
    }

    LL node=0;
    LL now=1;
    REP(i,n+1){
        node += now;
        now-=a[i];
        if(i!=n){
            if(now<=0){
                cout<<-1<<endl;
                return;
            }
            if(now>c[i+1]){
                cout<<-1<<endl;
                return;
            }
            now = min(now*2, c[i+1]);
        }
    }
    if(now != 0){
        cout<<-1<<endl;
    }
    else{
        cout<<node<<endl;
    }


}

int main()
{
	cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(15);
	Main();
    return 0;
}