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;
#define ll long long
#define ull unsigned long long
#define i128 __int128
#define mp make_pair
#define ld long double
typedef pair<int, int> pii;
typedef priority_queue<int, vector<int>, greater<int> > small_heap;
typedef priority_queue<int> big_heap;
const int N = 1e6 + 100;
int T;
ll a[N], r[N], l[N];

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i <= n; i++)scanf("%d", &a[i]);
    if (n == 0) {
        if (a[0] == 1)puts("1");
        else puts("-1");
        return 0;
    }
    for (int i = n; i >= 0; i--) {
        l[i] = a[i] + (l[i + 1] + 1) / 2;
        r[i] = a[i] + r[i + 1];
    }
    if (l[0] != 1) {
        puts("-1");
        return 0;
    }
    ll ans = 1, cur = 1;
    for (int i = 1; i <= n; i++) {
        cur -= a[i - 1];
        cur = min(cur * 2, r[i]);
        ans += cur;
    }
    printf("%lld\n", ans);
    return 0;
}