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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<tuple>
#include<cmath>
#include<iomanip>


using namespace std;
typedef long long ll;
typedef vector<ll> v;
typedef vector<vector<ll>> vv;


#define MOD 1000000007
#define INF 1e15
#define MIN -1001001001
#define rep(i,k,N) for(int i=k;i<N;i++)
#define MP make_pair
#define MT make_tuple //tie,make_tuple は別物
#define PB push_back
#define PF push_front
#define all(x) (x).begin(),(x).end()


int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};


int main(){
    ll N;
    cin>>N;
    v A(N+1);
    rep(i,0,N+1)cin>>A[i];
    v max_l(N+1);
    bool ok = true;
    max_l[0] = 1-A[0];

    rep(i,1,N+1){
        max_l[i] = min(max_l[i-1]*2-A[i],ll(INF));
        if(max_l[i]<0)ok = false;
    }
    ll ans = A[N],now = A[N];
    for(ll i = N-1;i >=0; i--){
        if(now == 0)ok = false;
        if(now > max_l[i] * 2)ok = false;
        ll now_l = min(now,max_l[i]);
        now = now_l + A[i];
        ans += now;
    }
    if(ok == false)cout<<-1;
    else cout<<ans;

    return 0;
}