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 <algorithm>
#include <cstdio>
using namespace std;
template<class T> inline bool Cmin(T &x,T y) {return x>y?x=y,1:0;}
typedef long long ll;
const ll INF=1e18;
const int maxN=1e5+50;
int N;
int A[maxN];
ll sum[maxN];
int main() {
	scanf("%d",&N);
	for(int i=0;i<=N;++i) scanf("%d",&A[i]);
	for(int i=N;i>=0;--i) sum[i]=sum[i+1]+A[i];
	ll an=0,now=1;
	for(int i=0;i<=N;++i) {
		if(now<A[i]) {
			puts("-1"); return 0;
		}
		an+=now;
		now-=A[i];
		now<<=1;
		Cmin(now,sum[i+1]);
	}
	printf("%lld\n",an);
	return 0;
}