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;
vector<long long> A;
void calc() {
  A.push_back(4);
  A.push_back(7);
  for (int i = 0; i < A.size(); ++i) {
    if (A[i] >= 1e8) continue;
    A.push_back(A[i] * 10 + 4);
    A.push_back(A[i] * 10 + 7);
  }
}
int main() {
  calc();
  long long pl, pr, vl, vr, k;
  long long ans = 0;
  cin >> pl >> pr >> vl >> vr >> k;
  for (int i = 0; i + k - 1 < A.size(); ++i) {
    long long MinPl, MinPr, MaxVl, MaxVr, MinVl, MinVr, MaxPl, MaxPr;
    if (A[i] < pl || (i - 1 >= 0 && A[i - 1] >= pr))
      MinPr = MinPl = 0;
    else {
      MinPl = max(i - 1 >= 0 ? A[i - 1] + 1 : 0ll, pl);
      MinPr = min(pr, A[i]);
    }
    if (A[i + k - 1] > vr || (i + k < A.size() && A[i + k] <= vl))
      MaxVl = MaxVr = 0;
    else {
      MaxVl = max(vl, A[i + k - 1]);
      MaxVr = min(i + k < A.size() ? A[i + k] - 1 : 100000000000ll, vr);
    }
    if (A[i] < vl || (i - 1 >= 0 && A[i - 1] >= vr))
      MinVl = MinVr = 0;
    else {
      MinVr = min(vr, A[i]);
      MinVl = max(i - 1 >= 0 ? A[i - 1] + 1 : 0ll, vl);
    }
    if (A[i + k - 1] > pr || (i + k < A.size() && A[i + k] <= pl))
      MaxPr = MaxPl = 0;
    else {
      MaxPl = max(pl, A[i + k - 1]);
      MaxPr = min(i + k < A.size() ? A[i + k] - 1 : 1000000000000ll, pr);
    }
    if (MinPr && MinPl && MaxVr && MaxVl)
      ans += (MinPr - MinPl + 1) * (MaxVr - MaxVl + 1);
    if (MinVr && MinVl && MaxPr && MaxPl)
      ans += (MinVr - MinVl + 1) * (MaxPr - MaxPl + 1);
    if ((MinPr && MinPl && MaxVr && MaxVl && MinVr && MinVl && MaxPr &&
         MaxPl) &&
        MinPr == MaxVl && MinVr == MaxPl)
      --ans;
  }
  long double answer =
      (long double)ans / ((long double)(pr - pl + 1) * (vr - vl + 1));
  cout << fixed << setprecision(12) << answer;
  return 0;
}