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;
typedef long double ld;
typedef vector<ll> vi;
#define INF (1e9)

int main() {
  ll N, ans = 0;
  cin >> N;
  vi A(N + 1), B(N + 2);
  for (ll i = 0; i < N + 1; i++) {
    cin >> A[i];
  }
  for (ll i = N; i >= 0; i--) {
    B[i] = B[i + 1] + A[i];
  }

  ll x = 1;
  for (ll i = 0; i < N + 1; i++) {
    // cout << x << endl;
    if (x < A[i]) {
      ans = -1;
      break;
    }

    ans += A[i];
    x -= A[i];
    ans += min(B[i] - A[i], x);
    if (B[i] - A[i] >= x) x *= 2;
  }
  cout << ans << endl;
}

