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>

#define ln                '\n'
#define all(dat)           dat.begin(), dat.end()
#define loop(i, to)        for (int i = 0; i < to; ++i)
#define cont(i, to)        for (int i = 1; i <= to; ++i)
#define circ(i, fm, to)    for (int i = fm; i <= to; ++i)
#define foreach(i, dat)    for (__typeof(dat.begin()) i = dat.begin(); i != dat.end(); ++i)

typedef long long          num;

using namespace std;

const int nsz = int(1e5);
int n;
num a[nsz + 5], l[nsz + 5], r[nsz + 5], ans;

int main() {
	ios::sync_with_stdio(0);
	cin >> n;
	circ (i, 0, n) cin >> a[i];
	l[n] = r[n] = a[n];
	for (int i = n; i >= 1; --i) {
		r[i - 1] = r[i] + a[i - 1], l[i - 1] = (l[i] - 1) / 2 + 1 + a[i - 1];
	}
	num cl = 1, cr = 1;
	circ (i, 0, n) {
		cl = max(cl, l[i]), cr = min(cr, r[i]);
		ans += cr;
		if (cl > cr || cr < a[i]) { cout << -1 << ln; return 0; }
		cl -= a[i], cr -= a[i];
		cr <<= 1;
	}
	cout << ans << ln;
}