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;
int f[10000];
int next(int n) {
  if (n == 0) return 4;
  if (n % 10 == 4) return (n / 10) * 10 + 7;
  return next(n / 10) * 10 + 4;
}
int main() {
  int pl, pr, vl, vr, k, i, j;
  int l1, l2, r1, r2;
  long long tu = 0;
  scanf("%d", &pl);
  scanf("%d", &pr);
  scanf("%d", &vl);
  scanf("%d", &vr);
  scanf("%d", &k);
  long long mau = (long long)(pr - pl + 1) * (vr - vl + 1);
  f[0] = 0;
  for (i = 1; f[i - 1] <= 777777774; i++) f[i] = next(f[i - 1]);
  f[i] = 1000000001;
  for (j = 1; k + j - 1 < i; j++) {
    if (f[j - 1] >= pl)
      l1 = f[j - 1] + 1;
    else
      l1 = pl;
    if (f[j] < pr)
      r1 = f[j];
    else
      r1 = pr;
    if (f[k + j - 1] >= vl)
      l2 = f[k + j - 1];
    else
      l2 = vl;
    if (f[k + j] <= vr)
      r2 = f[k + j] - 1;
    else
      r2 = vr;
    tu += (long long)max(r1 - l1 + 1, 0) * max(r2 - l2 + 1, 0);
  }
  for (j = 1; k + j - 1 < i; j++) {
    if (f[j - 1] >= vl)
      l2 = f[j - 1] + 1;
    else
      l2 = vl;
    if (f[j] < vr)
      r2 = f[j];
    else
      r2 = vr;
    if (f[k + j - 1] >= pl)
      l1 = f[k + j - 1];
    else
      l1 = pl;
    if (f[k + j] <= pr)
      r1 = f[k + j] - 1;
    else
      r1 = pr;
    tu += (long long)max(r1 - l1 + 1, 0) * max(r2 - l2 + 1, 0);
  }
  if (k == 1) {
    for (j = 1; j < i; j++)
      if (f[j] <= pr && f[j] >= pl && f[j] <= vr && f[j] >= vl) tu--;
  }
  double kq = (double)tu / mau;
  printf("%.9lf", kq);
  return 0;
}