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;

int main(){
	long long N;
	cin >> N;
	vector<long long> A(N+1);
	vector<long long> lim(N+1);
	long long ans=0,loop;
	for(loop=0;loop<=N;loop++){
		cin >> A[loop];
	}
	long long tmp=0;
	for(loop=N;loop>-1;loop--){
		tmp+=A[loop];
		lim[loop]=tmp;
	}
	long long ps=1;
	for(loop=0;loop<=N;loop++){
		if(loop<N&&ps-1<A[loop]){
			ans=-1;
			break;
		}else{
			ans+=ps;
			if(loop<N){
				ps=min((ps-A[loop])*2,lim[loop+1]);
			}else if(ps!=lim[loop]){
				ans=-1;
			}
		}
	}
	cout << ans << endl;
	return 0;
}