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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class MalekDanceClub {
    public static long mod = 1000000007;
    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        String x = f.readLine();
        System.out.println(f(x));
    }
    
    public static long f(String x) {
        if (x.length() == 0)
            return 0;
        long ans = f(x.substring(1))*2;
        if (x.charAt(0) == '1') 
            ans += pow(2*x.length()-2);
        return ans % mod;
    }
    
    public static long pow(int n) {
        if (n == 0)
            return 1;
        long x = pow(n/2);
        x = x*x % mod;
        if (n % 2 == 1)
            x = x*2 % mod;
        return x;
    }
}