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>

typedef long long LL;
const int MN = 200005;

int N;
LL A[MN], B[MN], C[MN];

int main() {
	scanf("%d", &N);
	for (int i = 0; i <= N; ++i) scanf("%lld", &A[i]);
	if (!N) return puts(A[0] == 1 ? "1" : "-1"), 0;
	for (int i = N; i >= 0; --i) B[i] = B[i + 1] + A[i];
	C[0] = 1 - A[0];
	for (int i = 1; i <= N; ++i) {
		C[i] = 2 * C[i - 1] - A[i];
		if (C[i] < 0 || C[i] + A[i] == 0) return puts("-1"), 0;
		if (C[i] + A[i] > B[i]) C[i] = B[i] - A[i];
	}
	LL Ans = 0;
	for (int i = 0; i <= N; ++i) Ans += A[i] + C[i];
	printf("%lld\n", Ans);
	return 0;
}