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.IOException;
import java.io.InputStream;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.NoSuchElementException;
import java.util.PriorityQueue;
import java.util.concurrent.PriorityBlockingQueue;

public class Main {

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int n = sc.nextInt();
        long[] a = new long[n+1];
        long sum = 0;
        for (int i = 0; i < a.length; i++) {
            a[i] = sc.nextLong();
            sum += a[i];
        }

        if(a[0] == 1 && a.length == 1){
            System.out.println(1);
            return;
        }else if(a[0] > 0){
            System.out.println(-1);
            return;
        }

        long ans = 1;
        long cur = 1;
        for (int i = 1; i < a.length; i++) {
            if(cur == 0){
                System.out.println(-1);
                return;
            }
            if(sum >= cur * 2){
                cur *= 2;
            }else{
                cur = sum;
            }
            ans += cur;

            sum -= a[i];
            cur -= a[i];

            if(sum < 0 || cur < 0){
                System.out.println(-1);
                return;
            }
        }

        if(sum > 0){
            System.out.println(-1);
            return;
        }

        System.out.println(ans);


    }


    private static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;

        private boolean hasNextByte() {
            if (ptr < buflen) {
                return true;
            } else {
                ptr = 0;
                try {
                    buflen = in.read(buffer);
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (buflen <= 0) {
                    return false;
                }
            }
            return true;
        }

        private int readByte() {
            if (hasNextByte())
                return buffer[ptr++];
            else
                return -1;
        }

        private static boolean isPrintableChar(int c) {
            return 33 <= c && c <= 126;
        }

        public boolean hasNext() {
            while (hasNextByte() && !isPrintableChar(buffer[ptr]))
                ptr++;
            return hasNextByte();
        }

        public String next() {
            if (!hasNext())
                throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while (isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        public long nextLong() {
            if (!hasNext())
                throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) {
                throw new NumberFormatException();
            }
            while (true) {
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                } else if (b == -1 || !isPrintableChar(b)) {
                    return minus ? -n : n;
                } else {
                    throw new NumberFormatException();
                }
                b = readByte();
            }
        }

        public int nextInt() {
            long nl = nextLong();
            if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
                throw new NumberFormatException();
            return (int) nl;
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }

}
