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)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,N) for(int i=0;i<int(N);++i)
#define rep1(i,N) for(int i=1;i<int(N);++i)
#define all(a) (a).begin(),(a).end()
#define print(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<_<<", "; cerr<<"]"<<endl; }
#define printpair(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<"{"<<_.first<<","<<_.second<<"}"<<", "; cerr<<"]"<<endl; }
#define dump(x) cerr<<#x<<": "<<x<<endl;
#define bit(k) (1LL<<(k))
typedef long long ll;
typedef pair<int, int> i_i;
typedef pair<ll, ll> l_l;
template <class T> using vec = vector<T>;
template <class T> using vvec = vector<vec<T>>;
template< typename T1, typename T2 >
ostream &operator<<(ostream &os, const pair< T1, T2 >& p) {
  os << "{" <<p.first << ", " << p.second << "}";
  return os;
}
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

const int INF = (ll)1e9;
const ll INFLL = (ll)1e18+1;
const ll MOD = (ll)1e9+7;
const double PI = acos(-1.0);
/*
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const string dir = "DRUL";
*/
ll power_pow(ll a, ll n) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a;
        a = a * a;
        n >>= 1;
    }
    return res;
}
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    ll N;
    cin >> N;
    vec<ll> A(N+1);
    rep(i,N+1)cin >> A[i];
    vec<ll> MAXA(N+1);
    rep(i,N+1){
        ll tmp = power_pow(2, min(60,i));
        if(i == N)MAXA[i] = min(INFLL, tmp);
        else MAXA[i] = min(INFLL, tmp - 1);;
    }
    rep(i,N+1){
        if(A[i] > MAXA[i]){
            cout << -1 << endl;
            return 0;
        }
    }
    if(A[N] == 0){
        cout << -1 << endl;
        return 0;
    }
    vec<ll> cnt(N+1);
    vec<ll> MAX_CNT(N+1);
    MAX_CNT[0] = 1;
    rep(i,N){
        MAX_CNT[i+1] = min(INFLL, 2 * MAX_CNT[i] - A[i+1]);
    }
    cnt[N] = A[N];
    for(int i = N; i >= 1; i --){
        //MAX_CNT[i-1]と、cnt[i]の差
        cnt[i-1] = min(MAX_CNT[i-1], cnt[i]) + A[i-1];;
        ll d = cnt[i] - MAX_CNT[i-1];
        if(d >= cnt[i]){
            cout << -1 << endl;
            return 0;
        }
        /*
        if((cnt[i] + 1) / 2 + A[i-1] > MAX_CNT[i-1]){
            dump(i);
            cout << -1 << endl;
            return 0;
        }
        */
    }
    /*
    cerr << "AAA";
    print(A);
    print(MAXA);
    cerr <<"c";
    print(cnt);
    print(MAX_CNT);
    */
    ll ans = 0;
    rep(i,N+1)ans += cnt[i];
    cout << ans << endl;
}