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
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const double pi = acos(-1.0);
const int mod = 1000000007;
const int maxn = 2111111;
char s[5444];
long long POW(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  scanf("%s", s);
  int len = strlen(s);
  reverse(s, s + len);
  int i, delta = 0;
  for (i = len - 1; i >= 0; --i) {
    if (s[i] == '0')
      delta++;
    else
      break;
  }
  if (i < 0) {
    puts("0");
    return 0;
  }
  long long high = POW(2, i);
  long long t = POW(2, delta);
  int p = i;
  long long k, ans = high * high % mod;
  for (i = 0; i < p; ++i) {
    if (s[i] == '1') {
      k = POW(2, i);
      ans += k * high;
      ans %= mod;
    }
  }
  ans = ans * t % mod;
  cout << ans << endl;
  return 0;
}
