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<int> e;
int intersect(int la, int ra, int lb, int rb) {
  if (la > ra) return 0;
  if (lb > rb) return 0;
  int l = (la < lb) ? lb : la;
  int r = (ra < rb) ? ra : rb;
  if (l > r) return 0;
  return r - l + 1;
}
int main() {
  for (int j = 1; j <= 9; ++j) {
    int br = 1 << j;
    for (int k = 0; k < br; ++k) {
      int i = 0;
      for (int t = j - 1; t >= 0; --t)
        if (k & (1 << t))
          i = i * 10 + 7;
        else
          i = i * 10 + 4;
      e.push_back(i);
    }
  }
  int k, la, ra, lb, rb;
  long long z = 0;
  cin >> la >> ra >> lb >> rb >> k;
  if (k == 0) {
    e.insert(e.begin(), -1);
    e.push_back(1000000500);
  }
  for (int i = 0; i + k <= e.size(); ++i) {
    int l, r, p1, p2, p3, p4;
    if (k == 0) {
      if (i == e.size()) break;
      l = max(la, lb);
      r = min(ra, rb);
      z += (long long)(intersect(e[i] + 1, e[i + 1] - 1, la, ra)) *
           intersect(e[i] + 1, e[i + 1] - 1, lb, rb);
      continue;
    }
    if (i)
      l = e[i - 1] + 1;
    else
      l = -1;
    r = e[i];
    p1 = intersect(l, r, la, ra);
    p3 = intersect(l, r, lb, rb);
    l = e[i + k - 1];
    if (i + k == e.size())
      r = 1000000500;
    else
      r = e[i + k] - 1;
    p2 = intersect(l, r, lb, rb);
    p4 = intersect(l, r, la, ra);
    long long z1 = ((long long)p1) * p2;
    long long z2 = ((long long)p3) * p4;
    if (k == 1 && la <= e[i] && e[i] <= ra && lb <= e[i] && e[i] <= rb) --z2;
    z += z1 + z2;
  }
  cout << setprecision(15) << fixed
       << ((long double)z) / (rb - lb + 1) / (ra - la + 1) << endl;
  return 0;
}