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 rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int MAX_N = 100005;
const ll INF = 1e18;
ll A[MAX_N],L[MAX_N],R[MAX_N];

int main()
{
	int N;
	cin >> N;
	rep(i,N+1)
	{
		cin >> A[i];
	}

	L[N] = A[N];
	R[N] = A[N];
	for (int i = N-1; i >= 0; --i)
	{
		L[i] = A[i] + (L[i+1]+1) / 2;
		R[i] = A[i] + R[i+1];
	}
	if (L[0] > 1)
	{
		cout << -1 << endl;
		return 0;
	}
	ll ans = 0;
	ll temp = 1;
	rep(i,N+1)
	{
		ans += temp;
		temp -= A[i];
		temp = min(temp*2,R[i+1]);
	}
	cout << ans << endl;
	return 0;
}