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 A[1 << 11], size = 0;
void init() {
  A[size++] = 0;
  for (int n = 1; n <= 9; ++n) {
    for (int i = 0; i < (1 << n); ++i) {
      int t = 0;
      for (int j = 0; j < n; ++j) {
        t *= 10;
        if ((i >> (n - j - 1)) & 1)
          t += 7;
        else
          t += 4;
      }
      A[size++] = t;
    }
  }
  A[size++] = 1000000001;
}
int main() {
  init();
  int pl, pr, vl, vr, k;
  scanf("%d%d", &pl, &pr);
  scanf("%d%d", &vl, &vr);
  scanf("%d", &k);
  double tot = (double)(pr - pl + 1) * (vr - vl + 1);
  double cnt = 0;
  for (int i = 1; i + k < size; ++i) {
    if (pl <= A[i] && A[i + k - 1] <= vr) {
      int al = max(A[i - 1] + 1, pl), ar = min(A[i], pr);
      int bl = max(A[i + k - 1], vl), br = min(A[i + k] - 1, vr);
      if (al <= ar && bl <= br) {
        cnt += (double)(ar - al + 1) * (br - bl + 1);
      }
    }
    if (vl <= A[i] && A[i + k - 1] <= pr) {
      int al = max(A[i - 1] + 1, vl), ar = min(A[i], vr);
      int bl = max(A[i + k - 1], pl), br = min(A[i + k] - 1, pr);
      if (al <= ar && bl <= br) {
        cnt += (double)(ar - al + 1) * (br - bl + 1);
      }
    }
    if (k == 1 && pl <= A[i] && A[i] <= pr && vl <= A[i] && A[i] <= vr)
      cnt -= 1.0;
  }
  printf("%.9lf\n", cnt / tot);
  return 0;
}