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 <iostream>
#include <vector>
#include <string>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define int long long
 const int MAX_N=1e5+2;
int cnt[MAX_N],n,suf[MAX_N];
int ans=0;

void dfs(int x,int now)
{
	ans+=now;//cout<<now<<"\n";
	if(x==n)
    {
		if(cnt[x]!=now) ans=-1;
		return ;
	}
	now-=cnt[x];
	if(now<=0)
    {
        ans=-1;
        return ;
	}
	dfs(x+1,min(now*2,suf[x+1]));
}

signed main()
{
    cin>>n;
	for(int i=0;i<=n;i++)
    {
        cin>>cnt[i];
        suf[i]=cnt[i];
    }

	for(int i=n;i>=0;i--) suf[i]+=suf[i+1];
	dfs(0,1);
	cout<<ans;
}
