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 MX=1e6+5;

ll ara[MX], csm[MX];

int main(){
     ll i, j, k;
     ll n;
     cin >> n;
     for(i=0; i<=n; ++i) cin >> ara[i];
     for(i=n; i>=0; --i) csm[i]=csm[i+1]+ara[i];
     ll cur=1, ans=1;
     for(i=0; i<=n; ++i){
          cur-=ara[i];
          if(cur<0){
               cout << -1 << endl;
               return 0;
          }
          cur*=2LL;
          cur=min(cur, csm[i+1]);
          ans+=cur;
     }
     cout << ans << endl;
     return 0;
}
