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;

long long N , A[1000003] , num[1000003];

int main(){
	cin >> N; for(int i = 0 ; i <= N ; ++i) cin >> A[i];
	num[0] = 1 - A[0]; for(int i = 1 ; i <= N ; ++i) num[i] = (num[i - 1] > 1e18 ? num[i - 1] : num[i - 1] << 1) - A[i];
	for(int i = 0 ; i <= N ; ++i) if(num[i] < 0){puts("-1"); return 0;}
	if(!N){cout << 1; return 0;}
	int pos = 0; long long sum = 1; num[0] = 2;
	for(int i = 1 ; i <= N ; ++i){
		while(A[i]){
			long long t = min(A[i] , num[pos]); sum += (i - pos) * t;
			A[i] -= t; num[pos] -= t; if(!num[pos]) ++pos;
		}
	}
	cout << sum;
}
