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>
#define ll long long
#define ld long double
#define pii pair<int, int>
#define pll pair<ll, ll>
#define str string
#define fi first
#define se second
#define pb push_back
#define SET(a, b) memset(a, b, sizeof(a))
#define eps 1e-6
#define pi atan(1) * 4
#define mod 1000000007
#define inf 1000000000
#define llinf 1000000000000000000
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
#define FORl(i, a, b) for (ll i = (a); i <= (b); i++)
#define FORDl(i, a, b) for (ll i = (a); i >= (b); i--)
using namespace std;
ll num[100005], sum[100005];
int main () {
  int n;
  scanf("%d", &n);
  FOR(i, 0, n) {
    scanf("%lld", &num[i]);
  }
  if (n == 0) {
    if (num[0] != 1) printf("-1\n");
    else printf("1\n");
    return 0;
  }
  if (num[0] > 0 || num[n] == 0) {
    printf("-1\n"); return 0;
  }
  sum[n + 1] = 0;
  FORD(i, n, 0) {
    sum[i] = sum[i + 1] + num[i];
  }
  ll d = 1, cnt = 1;
  FOR(i, 0, n - 1) {
    d -= num[i];
    if (d < 1) {
      printf("-1\n"); return 0;
    }
    d = min(d * 2, sum[i + 1]);
    cnt += d;
  }
  printf("%lld\n", cnt);
  return 0;
}