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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long a, b = 1, k, v;
string s;
int main() {
  cin >> s;
  k = (long long)s.size();
  for (long long i = k - 1, j = 0; i > -1; i--, j++) {
    if (s[i] == '1') {
      v = 1;
      for (int y = 0; y < j; y++) {
        v *= 2;
        if (v > 1000000007) v -= 1000000007;
      }
      a += v;
      if (a > 1000000007) a -= 1000000007;
    }
  }
  for (long long i = 0; i < k - 1; i++) {
    b *= 2;
    if (b > 1000000007) b -= 1000000007;
  }
  cout << (a * b) % 1000000007;
  return 0;
}