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.io.PrintWriter;
import java.util.*;

public class Main {
	static final long MOD=1000000007;//998244353;
	public static void main(String[] args) {
		PrintWriter out = new PrintWriter(System.out);
		InputReader sc=new InputReader(System.in);
		int N=sc.nextInt();
		long[] A=sc.nextLongArray(N+1);
		long X=1;
		long count=0;
		long[] c=new long[N+1];
		for (int i = 0; i < A.length; i++){
			c[i]=X;
			if (X<A[i]) {
				System.out.println(-1);
				return;
			}
			X-=A[i];
			if (X>Long.MAX_VALUE/2) {
				X=Long.MAX_VALUE;
			}else {
				X*=2;
			}
		}
		c[N]=Math.min(c[N], A[N]);
		for (int i = N-1; i >= 0; i--) {
			c[i]=Math.min(c[i], A[i]+c[i+1]);
		}
		for (int i = 0; i < c.length; i++) {
			count+=c[i];
		}
		System.out.println(count);
 	}
	static class InputReader { 
		private InputStream in;
		private byte[] buffer = new byte[1024];
		private int curbuf;
		private int lenbuf;
		public InputReader(InputStream in) {
			this.in = in;
			this.curbuf = this.lenbuf = 0;
		}
 
		public boolean hasNextByte() {
			if (curbuf >= lenbuf) {
				curbuf = 0;
				try {
					lenbuf = in.read(buffer);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (lenbuf <= 0)
					return false;
			}
			return true;
		}
 
		private int readByte() {
			if (hasNextByte())
				return buffer[curbuf++];
			else
				return -1;
		}
 
		private boolean isSpaceChar(int c) {
			return !(c >= 33 && c <= 126);
		}
 
		private void skip() {
			while (hasNextByte() && isSpaceChar(buffer[curbuf]))
				curbuf++;
		}
 
		public boolean hasNext() {
			skip();
			return hasNextByte();
		}
 
		public String next() {
			if (!hasNext())
				throw new NoSuchElementException();
			StringBuilder sb = new StringBuilder();
			int b = readByte();
			while (!isSpaceChar(b)) {
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}
 
		public int nextInt() {
			if (!hasNext())
				throw new NoSuchElementException();
			int c = readByte();
			while (isSpaceChar(c))
				c = readByte();
			boolean minus = false;
			if (c == '-') {
				minus = true;
				c = readByte();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res = res * 10 + c - '0';
				c = readByte();
			} while (!isSpaceChar(c));
			return (minus) ? -res : res;
		}
 
		public long nextLong() {
			if (!hasNext())
				throw new NoSuchElementException();
			int c = readByte();
			while (isSpaceChar(c))
				c = readByte();
			boolean minus = false;
			if (c == '-') {
				minus = true;
				c = readByte();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res = res * 10 + c - '0';
				c = readByte();
			} while (!isSpaceChar(c));
			return (minus) ? -res : res;
		}
 
		public double nextDouble() {
			return Double.parseDouble(next());
		}
 
		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
 
		public long[] nextLongArray(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}
 
		public char[][] nextCharMap(int n, int m) {
			char[][] map = new char[n][m];
			for (int i = 0; i < n; i++)
				map[i] = next().toCharArray();
			return map;
		}
	}
}
