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 x first
#define y second
#define mp make_pair
#define pb push_back

typedef long long ll;

const ll MOD = 1e9+7;
const ll INF = 1e9+5;

int a[100005];
ll suff[100005];

int main() {
	int n;
	scanf("%d", &n);
	for (int i=0 ; i<=n ; i++) {
		scanf("%d", a+i);
	}
	for (int i=n ; i>=0 ; i--) suff[i] = suff[i+1]+a[i];
	ll num = 1;
	ll sol = 0;
	for (int i=0 ; i<=n ; i++) {
	//	cout <<"num = "<<num<<endl;
		sol += num;
		if (a[i]>num) {
			printf("-1\n");
			return 0;
		}
		num -= a[i];
		num = min(num*2, suff[i+1]);
	}
	printf("%lld\n", sol);
	return 0;
}