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 a[1029];
long long hay(int x, int y) {
  if (x > y) return 0;
  return y - x + 1;
}
int main() {
  a[0] = 0;
  int sz = 1, ini = 0;
  while (ini < sz) {
    long long x = a[ini++];
    if (x * 10 + 4 < 1000000000) a[sz++] = x * 10 + 4;
    if (x * 10 + 7 < 1000000000) a[sz++] = x * 10 + 7;
  }
  a[sz++] = 1000000000 + 1;
  long long p1, p2, v1, v2, k;
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  long long res = 0, tot = (p2 - p1 + 1) * (v2 - v1 + 1);
  for (int i = 0; i < sz - k - 1; i++) {
    res += hay(max(a[i] + 1, p1), min(a[i + 1], p2)) *
           hay(max(a[i + k], v1), min(a[i + k + 1] - 1, v2));
    res += hay(max(a[i] + 1, v1), min(a[i + 1], v2)) *
           hay(max(a[i + k], p1), min(a[i + k + 1] - 1, p2));
  }
  if (k == 1) {
    int lo = max(p1, v1), hi = min(p2, v2);
    for (int i = 1; i < sz - 1; i++)
      if (lo <= a[i] && a[i] <= hi) res--;
  }
  cout << fixed << setprecision(10) << (long double)res / tot << endl;
}