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>
#define rep(i, n) for (ll i = 0; i < (n); i++)
using ll = long long;
using namespace std;
template <typename T> using vec = std::vector<T>;

int main() {
    int N;
    cin >> N;
    vec<ll> A(N + 1), S(N + 2);
    rep(i, N + 1) cin >> A[i];

    for (int i = N; i >= 0; --i) {
        S[i] = S[i + 1] + A[i];
    }

    ll ans  = 0;
    ll next = 1;
    rep(i, N + 1) {
        ans += next;
        next -= A[i];
        if (next < 0) {
            cout << -1 << endl;
            return 0;
        }
        next = min(next * 2, S[i + 1]);
    }
    cout << ans << endl;
}
