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 n, m, i, j, k, x, y, tmp, ans;
char s[1005];
long long Multi(long long x) {
  long long ans = 1, a = 2;
  while (x > 0) {
    if (x & 1) ans = (ans * a) % 1000000007;
    x >>= 1;
    a = (a * a) % 1000000007;
  }
  return ans % 1000000007;
}
int main() {
  scanf("%s", &s);
  n = strlen(s);
  ans = 0;
  tmp = 0;
  for (i = n - 1; i >= 0; i--) {
    x = n - 1 - i;
    y = s[i] - '0';
    if (!y)
      continue;
    else
      tmp = (tmp + (Multi(x) % 1000000007)) % 1000000007;
  }
  x = tmp;
  y = x;
  if (x == 0) {
    printf("0\n");
    return 0;
  }
  {
    tmp = 0;
    for (i = n - 1; i >= 0; i--) {
      if (s[i] == '1') {
        tmp = n - 1 - i;
        break;
      }
    }
    y = Multi(n - 1) % 1000000007;
    ans = (y * x) % 1000000007;
    printf("%I64d\n", ans);
  }
  return 0;
}