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
// NOMURA2020C.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 1e5 + 200;

typedef long long ll;

int n, seq[MAX_N];
ll leaves, ans;

int main()
{
	scanf("%d", &n);
	for (int i = 0; i <= n; i++)
		scanf("%d", &seq[i]), leaves += seq[i];
	if (n == 0)
		if (seq[0] == 1)
			puts("1"), exit(0);
		else
			puts("-1"), exit(0);
	if (seq[0] != 0)
		puts("-1"), exit(0);
	ll bas = ans = 1;
	for (int i = 1; i <= n; i++)
	{
		bas <<= 1;
		if (seq[i] > bas)
			puts("-1"), exit(0);
		ans += min(bas, leaves);
		bas = min(bas, leaves) - seq[i];
		leaves -= seq[i];
	}
	printf("%lld\n", ans);
	return 0;
}