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;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;

int main()
{
ll n ;
cin >> n ;
vector<ll> a(n+1);
rep(i, n+1){
  cin >> a[i];
}
vector<ll> aa(n+1);
aa[n]=a[n];
rep(i,n)aa[n-i-1]=aa[n-i]+a[n-i-1];
aa[0]=1;

rep(i,n){
  if(aa[i]-a[i]<=0){cout << -1 << endl;return 0;}
  else if((aa[i]-a[i])*2<a[i+1]){cout << -1 << endl;return 0;}
  else aa[i+1]=min(aa[i+1],(aa[i]-a[i])*2);
}
if(a[n]!=aa[n]){cout << -1 << endl;return 0;}
for(auto r:aa)ans+=r;
cout << ans << endl;
return 0;
}