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;
using ll = long long;
using P = pair<ll,ll>;
#define rep(i,s,n) for(int i = s; i < (int)(n); i++)

int main() {
  int n;
  cin >> n;
  vector<ll>a(n+1);
  rep(i,0,n+1) cin >> a[i];
  vector<P>c(n+1);
  c[n].first = a[n];c[n].second = a[n];
  for(int i = n-1;i >= 0;i--){
    c[i].first = (c[i+1].first+1)/2+a[i];
    c[i].second = c[i+1].second+a[i];
  }
  if(c[0].first > 1){
    cout << -1 << endl;
    return 0;
  }
  ll ans = 1,v = 1;
  rep(i,1,n+1){
    ll t = min(2*v,c[i].second);
    ans += t;
    v = t-a[i];
  }
  cout << ans << endl;
}