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
import java.util.*;
import java.math.*;
import java.io.*;

public class Main
	{
	static long mod=1000000007;
	public static void main(String args[]) throws IOException
		{
		Scanner c=new Scanner(System.in);
		String s=c.nextLine();
		long ans=process(s);
		System.out.println(ans);
		}

	private static long process(String s)
		{
		//System.out.println(s);
		if(s.length()==0)
			return 0;
		int N=s.length();
		if(s.charAt(0)=='0')
			return (2*process(s.substring(1,s.length())))%mod;
		else
			{
			long ans1=modpow(2*N-2,mod);
			//System.out.println("ans1 --" +ans1);
			long ans2=process(s.substring(1,s.length()));
			ans2=(ans2*2)%mod;
			return (ans1+ans2)%mod;
			}
		}
	private static long modpow(int n, long mod)
		{
		long ans=1;
		while (n>0)
			{
			ans=ans*2;
			ans%=mod;
			n--;
			}
		return ans;
		}
	}