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;
long long int arr[1024], i, j, k, l, pin, pfi, vin, vfi, in;
long double temp, res;
double resf;
long long int cuen(long long int ina, long long int fia, long long int inb,
                   long long int fib) {
  if ((inb >= ina) && (fib <= fia)) {
    return fib - inb + 1;
  }
  if ((ina >= inb) && (fia <= fib)) {
    return fia - ina + 1;
  }
  if ((ina <= fib) && (fib <= fia)) {
    return fib - ina + 1;
  }
  if ((ina <= inb) && (inb <= fia)) {
    return fia - inb + 1;
  }
  return 0;
}
int main() {
  arr[0] = 0;
  arr[1] = 4;
  arr[2] = 7;
  in = 1;
  k = 3;
  l = 3;
  for (i = 2; i <= 9; i++) {
    for (j = in; j < k; j++) {
      arr[l] = arr[j] * 10 + 4;
      arr[l + 1] = arr[j] * 10 + 7;
      l += 2;
    }
    in = k;
    k = l;
  }
  arr[l] = 1000000002;
  cin >> pin >> pfi >> vin >> vfi >> k;
  for (i = 1; i <= 1022 - k + 1; i++) {
    int inix = arr[i], finx = arr[i + k - 1];
    int iniy, finy;
    iniy = arr[i - 1] + 1;
    finy = arr[i + k] - 1;
    temp += (cuen(iniy, inix, vin, vfi) * cuen(finx, finy, pin, pfi));
    temp += (cuen(iniy, inix, pin, pfi) * cuen(finx, finy, vin, vfi));
    if (k == 1)
      if ((cuen(vin, vfi, arr[i], arr[i]) > 0) &&
          (cuen(pin, pfi, arr[i], arr[i]) > 0)) {
        temp--;
      }
  }
  res = ((vfi - vin + 1) * (pfi - pin + 1));
  res = temp / res;
  resf = res;
  printf("%.18f", resf);
}