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>

#define ll long long
using namespace std;

ll solve()
{
	int N;
    cin>>N;
	++N;
	vector<ll> A(N);
    for(ll &a : A)
        cin>>a;
	vector<ll> B(N);
	B[N-1] = A[N-1];
	for(int i=N-2; i>=0; --i)
	{
		B[i]=B[i+1]+A[i];
	}
	B[0]=1;
	if(B[0]<A[0])
		return -1;
	for(int i=1; i<N;++i)
	{
		ll prev = B[i-1]-A[i-1];
		B[i]=min(B[i],prev*2);
		if(B[i]<A[i])
			return -1;
	}
	ll ans = 0;
	for(auto &a: B)
		ans+=a;
	return ans;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
    

    cout<<solve()<<endl;
	return 0;
}