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;
int n,i,a[100100];
long long mn[100100],mx[100100],r;
int main() {
  scanf("%d",&n);
  for (i=0; i<=n; i++) scanf("%d",&a[i]);
  for (i=n; i>=0; i--) {
    mn[i]=(mn[i+1]+1)/2+a[i];
    if (i<30 && mn[i]>(1<<i)) break;
    mx[i]=mx[i+1]+a[i];
    if (i<30 && mx[i]>(1<<i)) mx[i]=(1<<i);
  }
  if (i>=0) { puts("-1"); return 0; }
  for (i=0; i<=n; i++) {
    mx[i+1]=min(mx[i+1],(mx[i]-a[i])*2);
    if (mx[i]<mn[i]) break;
    r+=mx[i];
  }
  if (i<=n) puts("-1"); else printf("%lld\n",r);
  return 0;
}
