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.util.*;
import java.io.*;
import static java.lang.Math.*;

public class Main {
    public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int[] a = new int[n+1];
		for(int i = 0; i < n+1; i++){
		    a[i] = sc.nextInt();
		}
		
		// 計算
		long result = 0;
		long[] upper = new long[n+1];
		long[] min = new long[n+1];
		for(int i = 0; i < n+1; i++){
		    if(i == 0){
		        upper[n-i] = a[n-i];
		        min[n-i] = a[n-i];
		    }else{
		        upper[n-i] = upper[n-i+1] + a[n-i];
		        min[n-i] = (min[n-i+1]+1)/2 + a[n-i];
		    }
		}
		for(int i = 0; i < n+1; i++){
		    if(i == 0){
		        if(min[i] > 1){
		            result = -1;
		            break;
		        }
		        upper[i] = 1;
		        result += upper[i];
		    }else{
		        upper[i] = min(upper[i], (upper[i-1]-a[i-1])*2);
		        result += upper[i];
		    }
		}
		
		// 出力
		System.out.println(result);
	}

	public static class Scanner {
		private BufferedReader br; private StringTokenizer tok;
		public Scanner(InputStream is) throws IOException { br = new BufferedReader(new InputStreamReader(is));}
		private void getLine() throws IOException { while(!hasNext()) tok = new StringTokenizer(br.readLine());}
		private boolean hasNext() { return tok != null && tok.hasMoreTokens();}
		public String next() throws IOException { getLine(); return tok.nextToken();}
		public int nextInt() throws IOException { return Integer.parseInt(next());}
		public long nextLong() throws IOException { return Long.parseLong(next());}
		public double nextDouble() throws IOException { return Double.parseDouble(next());}
	}
}
