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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <math.h>
#include <queue>
#include <string>
#define ll long long
using namespace std;
 
int main() {
  vector<ll> a;
  vector<ll> ps;
  int n; cin >> n;
  ll ps_sum = 0;
  for(int i=0; i<=n; i++){
    ll x; cin >> x;
    a.push_back(x);
    ps.push_back(ps_sum + x);
    ps_sum += x;
  }

  ll sum = 0;
  ll na = 1;
  for(int i=0; i<=n; i++){
    if(a[i]>na || (a[i]==na && i<n)){
      sum = -1;
      break;
    }
    na = min(na-a[i], ps[n]-ps[i]);
    sum += na + a[i];
    na *= 2;
  }

  cout << sum;
}