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
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        MyScanner in = new MyScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        C solver = new C();
        solver.solve(1, in, out);
        out.close();
    }

    static class C {
        public void solve(int testNumber, MyScanner in, PrintWriter out) {
            int N = in.Int();
            int[] A = in.Int(N + 1);
            if (N == 0) {
                if (A[0] <= 1) {
                    out.println(1);
                } else {
                    out.println(-1);
                }
                return;
            }
            if (N > 0 && A[0] > 0) {
                out.println(-1);
                return;
            }

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

            long ans = 1;
            long cur = 1;
            for (int i = 1; i < N; i++) {
                long temp = cur * 2 - A[i];
                if (temp <= 0) {
                    out.println(-1);
                    return;
                }
                cur = Math.min(temp, B[i + 1]);
                ans += cur + A[i];
            }
            if (cur > A[N] || cur * 2 < A[N]) {
                out.println(-1);
            } else {
                out.println(ans + A[N]);
            }
        }

    }

    static class MyScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public MyScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    String rl = in.readLine();
                    if (rl == null) {
                        return null;
                    }
                    st = new StringTokenizer(rl);
                } catch (IOException e) {
                    return null;
                }
            }
            return st.nextToken();
        }

        public int Int() {
            return Integer.parseInt(next());
        }

        public int[] Int(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = Int();
            }
            return a;
        }

    }
}

