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;

ll solve() {
    ll N;
    cin >> N;
    vector<ll> A(N+1);
    for ( int i = 0; i < N+1; i++ ) {
        cin >> A[i];
    }

    ll w = 0;
    vector<ll> mf(N+1);
    for ( int i = N; i >= 1; i-- ) {
        w += A[i];
        mf[i-1] = w;
    }

    ll v = 1, c = 0;
    for ( int i = 0; i < N+1; i++ ) {
        if ( A[i] > v ) return -1;
        c += A[i];
        v -= A[i];
        if ( v > mf[i] ) v = mf[i];
        c += v;
        v *= 2;
    }

    return c;
}

int main() {
    auto ans = solve();
    cout << ans << "\n";
    return 0;
}