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;
const int OO = 0x3f3f3f3f, N = 1e4 + 5, mod = 1e9 + 7;
string str;
int fast_power(int b, int p) {
  int res = 1;
  while (p) {
    if (p % 2) res = (1ll * res * b) % mod;
    b = (1ll * b * b) % mod;
    p /= 2;
  }
  return res;
}
int solve(int idx, int sz) {
  if (idx == str.size()) return 0;
  return ((2ll * solve(idx + 1, sz - 1)) % mod +
          ((str[idx] == '1') ? fast_power(2, 2 * sz) : 0)) %
         mod;
}
int main() {
  cin >> str;
  cout << solve(0, str.size() - 1);
  return 0;
}