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 Cross(long long L1, long long R1, long long L2, long long R2) {
  if (max(L1, L2) > min(R1, R2)) return 0;
  return min(R1, R2) - max(L1, L2) + 1;
}
int main() {
  int Total = 0, Lucky[1024];
  for (int i = 1; i <= 9; i++)
    for (int j = 0; j < (1 << i); j++) {
      int S = 0;
      for (int k = 0; k < i; k++) S = S * 10 + ((j & (1 << k)) ? 4 : 7);
      Lucky[Total++] = S;
    }
  Lucky[Total++] = 0;
  Lucky[Total++] = 1000000001;
  sort(Lucky, Lucky + Total);
  int L1, R1, L2, R2, K;
  cin >> L1 >> R1 >> L2 >> R2 >> K;
  long long Count = 0;
  for (int Left = 1, Right = K; Right < Total - 1; Left++, Right++) {
    Count += Cross(L1, R1, Lucky[Left - 1] + 1, Lucky[Left]) *
             Cross(L2, R2, Lucky[Right], Lucky[Right + 1] - 1);
    Count += Cross(L2, R2, Lucky[Left - 1] + 1, Lucky[Left]) *
             Cross(L1, R1, Lucky[Right], Lucky[Right + 1] - 1);
  }
  if (K == 1)
    for (int i = 1; i < Total - 1; i++)
      if (Lucky[i] >= max(L1, L2) && Lucky[i] <= min(R1, R2)) Count--;
  double Ans = double(Count) / (R1 - L1 + 1) / (R2 - L2 + 1);
  printf("%0.10lf\n", Ans);
  return 0;
}