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.

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.

Examples

Input

3
0 1 1 2


Output

7


Input

4
0 0 1 0 2


Output

10


Input

2
0 3 1


Output

-1


Input

1
1 1


Output

-1


Input

10
0 0 1 1 2 3 5 8 13 21 34


Output

264
#include <cstdio>
#include <algorithm>
using namespace std;
const long long INF = 1LL << 60;
int main() {
	int N;
	scanf("%d", &N);
	long long A[N + 1], mx[N + 1], ans = 0;
	for (int i = 0; i <= N; ++i) scanf("%lld", &A[i]);
	if (A[0] <= 1) {
		mx[0] = 1 - A[0];
		for (int i = 1; i <= N; ++i) {
			mx[i] = min(mx[i - 1] * 2, i < 60 ? 1LL << i : INF) - A[i];
			if (mx[i] < 0) {
				printf("-1");
				return 0;
			}
		}
		mx[N] = 0;
		for (int i = N - 1; i >= 0; --i) mx[i] = min(mx[i], mx[i + 1] + A[i + 1]);
		for (int i = 0; i <= N; ++i) ans += mx[i] + A[i];
		printf("%lld", ans);
	} else printf("-1");
	return 0;
}