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 l1, r1, l2, r2;
int k;
int mass[7027];
int tem[7027];
int coun;
void mergesort(int *a, int *b, int n) {
  int i, i0, i1, iRight, iEnd, width, j;
  for (width = 1; width < n; width *= 2)
    for (i = 0; i < n; i += width * 2) {
      i0 = i;
      i1 = iRight = min(i + width, n);
      iEnd = min(i + width * 2, n);
      for (j = i; j < iEnd; j++)
        if (i0 < iRight && (i1 == iEnd || a[i0] < a[i1]))
          b[j] = a[i0++];
        else
          b[j] = a[i1++];
      for (j = i; j < iEnd; j++) a[j] = b[j];
    }
}
int d[10];
void dfs(int x, int res) {
  mass[coun++] = res + d[x] * 4;
  mass[coun++] = res + d[x] * 7;
  if (x == 8) return;
  dfs(x + 1, res + d[x] * 4);
  dfs(x + 1, res + d[x] * 7);
}
int main() {
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  long long ans = 0;
  int i, j;
  d[0] = 1;
  for (i = 1; i < 10; i++) d[i] = d[i - 1] * 10;
  dfs(0, 0);
  mergesort(mass, tem, coun);
  for (i = coun; i > 0; i--) mass[i] = mass[i - 1];
  mass[coun + 1] = 2000000000;
  mass[0] = -2000000000;
  for (i = 1; i + k <= coun + 1; i++) {
    if (mass[i] >= l1 && mass[i - 1] < r1)
      if (mass[i + k] > l2 && mass[i + k - 1] <= r2)
        ans += (long long)(min(r1, mass[i]) - max(l1, mass[i - 1] + 1) + 1) *
               (min(r2, mass[i + k] - 1) - max(l2, mass[i + k - 1]) + 1);
    if (mass[i] >= l2 && mass[i - 1] < r2)
      if (mass[i + k] > l1 && mass[i + k - 1] <= r1)
        ans += (long long)(min(r2, mass[i]) - max(l2, mass[i - 1] + 1) + 1) *
               (min(r1, mass[i + k] - 1) - max(l1, mass[i + k - 1]) + 1);
  }
  if (k == 1) {
    for (i = 1; i <= coun; i++)
      if (l1 <= mass[i] && mass[i] <= r1 && l2 <= mass[i] && mass[i] <= r2)
        ans--;
  }
  long long del = (long long)(r1 - l1 + 1) * (r2 - l2 + 1);
  cout << setprecision(20) << (long double)ans / (long double)del << endl;
  return 0;
}