Problem: As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.

One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.

The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.

You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).

Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».

Input

The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).

This number may contain leading zeros.

Output

Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).

Examples

Input

11


Output

6


Input

01


Output

2


Input

1


Output

1

Solution (in JAVA):

Solution:
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
 *
 * @author Nafiur Rahman Khadem Shafin
 */
public class Main {
	public static void main (String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader (inputStream);
		PrintWriter out = new PrintWriter (outputStream);
		ProblemAMalekDanceClub solver = new ProblemAMalekDanceClub ();
		solver.solve (1, in, out);
		out.close ();
	}
	
	static class ProblemAMalekDanceClub {
		private static long modpowx (int x, long n) {
			long res = 1, os = x;
			while (n>0) {
				if ((n&1) != 0) res = (res*os)%InputReader.inf;
				os = (os*os)%InputReader.inf;
				n >>= 1;
			}
			return res;
		}
		
		private static long func (int i, int n) {
			long os = modpowx (2, i);
			os %= InputReader.inf;
			os *= modpowx (2, n-1-i);
			os %= InputReader.inf;
			os *= modpowx (2, n-1-i);
			os %= InputReader.inf;
			return os;
		}
		
		public void solve (int testNumber, InputReader in, PrintWriter out) {
			String x = in.next ();
			char[] arr = x.toCharArray ();
			long res = 0;
			for (int i = 0; i<arr.length; i++) {
				if (arr[i] == '1') {
					res += func (i, arr.length);
					res %= InputReader.inf;
				}
			}
			out.println (res);
		}
		
	}
	
	static class InputReader {
		private BufferedReader reader;
		private StringTokenizer tokenizer;
		public static final int inf = 1000000007;
		
		public InputReader (InputStream stream) {
			reader = new BufferedReader (new InputStreamReader (stream));
			tokenizer = null;
		}
		
		public String next () {
			while (tokenizer == null || !tokenizer.hasMoreTokens ()) {
				try {
					String str;
					if ((str = reader.readLine ()) != null) tokenizer = new StringTokenizer (str);
					else return null;//to detect eof
				} catch (IOException e) {
					throw new RuntimeException (e);//😂😁😆😄😜🙂😞😊😍🤔❤💔😕😠😑😮🤐😢😱👍👎👌👏😎🤦
				}
			}
			return tokenizer.nextToken ();
		}
		
	}
}