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;
typedef long long ll;
const int sz = 1e5+5;

int a[sz];

int main()
{
   int n;
   ll tot = 0;
   scanf("%d", &n);
   for(int i=0; i<=n; i++)
      scanf("%d", &a[i]), tot += a[i];

   ll ans = 0, x = 1;
   for(int i=0; i<=n; i++) {
      if(x < a[i]) {
         puts("-1");
         exit(0);
      }
      ans += x;
      x -= a[i];
      tot -= a[i];
      x = min(x << 1, tot);
   }

   if(!n && !a[0]) ans = 0;
   printf("%lld\n", ans);
}
