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 fi first
#define se second
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define endl '\n'

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, int> pdi;

const ll INF = 2e18;
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;

ll N, A[MAXN], L[MAXN], R[MAXN], ans;

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	cin >> N;
	for(int i=0; i<=N; i++) cin >> A[i];
	if(N == 0) {
		if(A[0] == 1) cout << 1;
		else if(A[0] == 0) cout << 0;
		else cout << -1;
		return 0;
	}
	L[N] = R[N] = A[N];
	for(int i=N-1; i>=0; i--) {
		L[i] = (L[i+1]+1)/2 + A[i];
		R[i] = R[i+1] + A[i];
		if(i<=60) R[i] = min(1LL << i, R[i]);
		if(L[i] > R[i]) {
			cout << -1;
			return 0;
		}
	}
	ll bef = 1; ans = 1;
	for(int i=1; i<=N; i++) {
		ll t = min(R[i], (bef-A[i-1])*2);
		if(t < L[i]) {
			cout << -1;
			return 0;
		}
		bef = t;
		ans += t;
	}
	cout << ans;
}