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 rep(i,n) for(int i=0;i<(n);++i)
#define rrep(i,n) for(int i=1;i<(n);++i)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define maxs(a, b) a = max(a, b)
#define mins(a, b) a = min(a, b)
using namespace std;
typedef long long ll;
typedef pair<double, int> P;
const ll linf = (1ll << 61);
const int inf = 1001001001;
const int mod = 1000000007;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	vector<ll> a(n + 1);
	rep(i, n + 1) cin >> a[i];
	if (n != 0 && a[0] != 0) {cout << -1 << endl;return 0;}
	if (n == 0 && a[0] != 1) {cout << -1 << endl;return 0;}

	vector<ll> mx(n + 1);
	ll now = 0;
	for (int i = n; i >= 0; --i) {
		now += a[i];
		mx[i] += now;
	}
	ll ans = 1;
	now = 1;
	for (int i = 1; i <= n; ++i) {
		now *= 2;
		if (now < a[i]) {cout << -1 << endl;return 0;}
		ans += min(mx[i], now);
		now = min(now - a[i], mx[i]);
	}
	cout << ans << endl;
 	return 0;
}