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;
using ll=long long;
constexpr ll mod=1000000007;
constexpr ll mod998=998244353;

int N,A[100001];
ll M[100002];
ll res;


int main(){
	cin>>N;
	for(int i=0;i<=N;i++){
		cin>>A[i];
		M[i]=A[i];
	}
	for(int i=N-1;i>=0;i--){
		M[i]+=M[i+1];
	}
	ll t=1;
	for(int i=0;i<=N;i++){
		t=min(t,M[i]);
		res+=t;
		t-=A[i];
		if(t<0){
			puts("-1");
			return 0;
		}
		t*=2;
	}
	cout<<res<<endl;
}
