Problem: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.

Input

The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).

Output

On the single line print the result with an absolute error of no more than 10 - 9.

Examples

Input

1 10 1 10 2


Output

0.320000000000


Input

5 6 8 10 1


Output

1.000000000000

Note

Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>

In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.

In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void ggg(vector<int>& allLucky, int len, int v = 0) {
  if (len == 0)
    allLucky.push_back(v);
  else {
    ggg(allLucky, len - 1, 10 * v + 4);
    ggg(allLucky, len - 1, 10 * v + 7);
  }
}
int main(int argc, char* argv[]) {
  vector<int> allLucky;
  for (int len = 1; len < 10; len++) ggg(allLucky, len);
  int p_l, p_r, v_l, v_r;
  int num;
  cin >> p_l >> p_r >> v_l >> v_r >> num;
  int i_p_l =
      upper_bound(allLucky.begin(), allLucky.end(), p_l) - allLucky.begin();
  int i_p_r =
      upper_bound(allLucky.begin(), allLucky.end(), p_r) - allLucky.begin();
  int i_v_l =
      upper_bound(allLucky.begin(), allLucky.end(), v_l) - allLucky.begin();
  int i_v_r =
      upper_bound(allLucky.begin(), allLucky.end(), v_r) - allLucky.begin();
  vector<int> ppp, vvv;
  ppp.push_back(p_l);
  ppp.insert(ppp.end(), allLucky.begin() + i_p_l, allLucky.begin() + i_p_r);
  if (p_r > *(ppp.rbegin())) ppp.push_back(p_r);
  vvv.push_back(v_l);
  vvv.insert(vvv.end(), allLucky.begin() + i_v_l, allLucky.begin() + i_v_r);
  if (v_r > *(vvv.rbegin())) vvv.push_back(v_r);
  long long sum_ok = 0LL;
  double sum_all = (p_r - p_l + 1.0) * (v_r - v_l + 1.0);
  vector<int> num_less_p, num_l_eq_p, num_less_v, num_l_eq_v;
  for (int v = 0; v < vvv.size(); v++) {
    num_less_v.push_back(
        (lower_bound(allLucky.begin(), allLucky.end(), vvv[v]) + 1) -
        allLucky.begin());
    num_l_eq_v.push_back(
        (upper_bound(allLucky.begin(), allLucky.end(), vvv[v]) + 1) -
        allLucky.begin());
  }
  for (int p = 0; p < ppp.size(); p++) {
    num_less_p.push_back(
        (lower_bound(allLucky.begin(), allLucky.end(), ppp[p]) + 1) -
        allLucky.begin());
    num_l_eq_p.push_back(
        (upper_bound(allLucky.begin(), allLucky.end(), ppp[p]) + 1) -
        allLucky.begin());
  }
  for (unsigned v = 0; v < vvv.size(); v++)
    for (unsigned p = 0; p < ppp.size(); p++) {
      if (v > 0 && p > 0)
        if (abs(num_less_p[p] - num_less_v[v]) == num)
          sum_ok += (long long)(ppp[p] - ppp[p - 1] - 1) *
                    (long long)(vvv[v] - vvv[v - 1] - 1);
      if (v > 0)
        if (num_l_eq_p[p] - num_less_v[v] == num ||
            num_less_v[v] - num_less_p[p] == num)
          sum_ok += vvv[v] - vvv[v - 1] - 1;
      if (p > 0)
        if (num_less_p[p] - num_less_v[v] == num ||
            num_l_eq_v[v] - num_less_p[p] == num)
          sum_ok += ppp[p] - ppp[p - 1] - 1;
      if (num_l_eq_p[p] - num_less_v[v] == num ||
          num_l_eq_v[v] - num_less_p[p] == num)
        sum_ok += 1LL;
    };
  cout.precision(9);
  cout << (double)(sum_ok) / (double)(sum_all);
  return 0;
}