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 <iostream>

using namespace std;

typedef long long ll;

int n;
ll a[101010];
ll s[101010];

void rip() {
	cout << "-1\n";
	exit(0);
}

int main() {
	cin >> n;
	for (int i = 0; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = n; i >= 0; i--) s[i] = s[i+1]+a[i];
	ll c = 1;
	ll u = 0;
	for (int i = 0; i <= n; i++) {
		u += c;
		if (a[i] > c) rip();
		c -= a[i];
		if (i == n) break;
		if (c == 0 || c > s[i+1]) rip;
		c = min(2*c,s[i+1]);
	}
	if (c != 0) rip();
	cout << u << "\n";
}
