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<bits/stdc++.h>
using namespace std;
#define ll long long
const int INF = 0x3f3f3f3f;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 10;
const int mod = 998244353;
int n;
ll lea[N], sum[N], a[N], ans = 1;
int main() {
	scanf("%d", &n);
	for (int i = 0; i <= n; i++)scanf("%lld", &lea[i]);
	if (lea[0] > 1) { puts("-1"); return 0; }
	for (int i = n; i >= 0; i--)sum[i] = sum[i + 1] + lea[i];
	a[0] = 1;
	for (int i = 1; i <= n; i++) {
		if (lea[i] > 2 * (a[i - 1] - lea[i - 1])) { puts("-1"); return 0; }
		a[i] = min(2 * (a[i - 1] - lea[i - 1]), sum[i]);
		ans += a[i];
	}
	printf("%lld\n", ans);
	return 0;
}