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 <algorithm>
#include <iostream>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cstdio>
#include <bitset>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <map>
#include <set>

#define fi first
#define se second
#define db double
#define U unsigned
#define P std::pair<int,int>
#define LL long long
#define pb push_back
#define MP std::make_pair
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define FOR(i,a,b) for(int i = a;i <= b;++i)
#define ROF(i,a,b) for(int i = a;i >= b;--i)
#define DEBUG(x) std::cerr << #x << '=' << x << std::endl
#define int LL
const int MAXN = 5e5 + 5;
const int MAX = 1e18;
int n,a[MAXN],mx[MAXN],ans[MAXN];

signed main(){
    scanf("%lld",&n);LL deg0 = 0;
    FOR(i,0,n) scanf("%lld",a+i);
    if(n == 0 && a[0]!= 1){
        puts("-1");
        return 0;
    }
    if(n != 0 && a[0] != 0){
        puts("-1");
        return 0;
    }
    mx[0] = 1;
    FOR(i,1,n){
        mx[i] = (mx[i-1]-a[i-1])*2;
        if(mx[i] < a[i]){
            puts("-1");
            return 0;
        }
        if(mx[i] <= 0){
            puts("-1");
            return 0;
        }
        mx[i] = std::min(mx[i],MAX);
    }
    int res = 0;
    ROF(i,n,0){
        ans[i] = a[i]+ans[i+1];
        ans[i] = std::min(ans[i],mx[i]);
        res += ans[i];
    }
    printf("%lld\n",res);
    return 0;
}