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 <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <numeric>
#include <functional>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <sstream>
#include <string>
#define _repargs(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _repargs(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define mod 1000000007
#define inf 2000000007
#define mp make_pair
#define pb push_back
typedef long long ll;
using namespace std;
template <typename T>
inline void output(T a, int p = 0) {
    if(p) cout << fixed << setprecision(p)  << a << "\n";
    else cout << a << "\n";
}
// end of template

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    // source code
    int N;
    cin >> N;
    vector<ll> A(N + 1);
    rep(i, N + 1) cin >> A[i];
    reverse(all(A));
    if(N == 0 && A[0] != 1) {
        output(-1);
        return 0;
    }
    if(N >= 1 && A[N] != 0) {
        output(-1);
        return 0;
    }
    
    vector<pair<ll, ll>> mm(N + 1);
    rep(i, N + 1) {
        if(i == 0) mm[i] = mp(A[i], A[i]);
        else {
            mm[i] = mp(A[i] + (mm[i - 1].first + 1) / 2, A[i] + mm[i - 1].second);
        }
    }
    
    ll cur = 1;
    bool ok = true;
    vector<ll> B(N + 1);
    for(int i = N; i >= 0; i--) {
        if(i == N) {
            B[i] = cur;
            cur -= A[i];
        }
        else {
            if(mm[i].first > cur * 2 || cur > mm[i].second) {
                ok = false;
                break;
            }
            cur *= 2;
            cur = min(cur, mm[i].second);
            B[i] = cur;
            cur -= A[i];
        }
    }
    
    if(!ok) {
        output(-1);
        return 0;
    }
    ll ans = 0;
    rep(i, N + 1) ans += B[i];
    output(ans);
    
    
    
    return 0;
}