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;
typedef long long LL;
const int N=1001001;
const LL INF=1e15;
void gg(){
	puts("-1");
	exit(0);
}
int n;
int a[N];
LL suf[N];
int main(){
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=0;i<=n;i++){
		cin>>a[i];
	}
	/**
	 * t=1;
	 * {
	 * ans+=t;
	 * t-=a[i];
	 * t=t...t*2;
	 * }
	 * assert(t==0);
	 */
	for(int i=n;i>=0;i--){
		suf[i]=suf[i+1]+a[i];
	}
	LL t=1;
	LL ans=0;
	for(int i=0;i<=n;i++){
		ans+=t;
		t-=a[i];
		if(t<0)gg();
		LL s=suf[i+1];
		assert(s>=t);
		t*=2;
		t=min(t,suf[i+1]);
	}
	cout<<ans<<endl;
	return 0;
}