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 res7 = 10000000, res4 = 10000001;
int vl, vr, pl, pr;
double getp(int la, int na, int a, int aa) {
  double ret = 0;
  if (vl <= a && pr >= aa && min(a, vr) >= max(la, vl) &&
      min(na, pr) >= max(aa, pl)) {
    ret += (double)(min(a, vr) - max(la, vl) + 1) / (vr - vl + 1) *
           (double)(min(na, pr) - max(aa, pl) + 1) / (pr - pl + 1);
  }
  return ret;
}
int main() {
  vector<int> a;
  for (int n = 1; n <= 9; ++n) {
    for (int I = 0; I < (1 << n); ++I) {
      int x = 0;
      for (int i = 0; i < n; ++i) {
        if (I & (1 << i))
          x = x * 10 + 4;
        else
          x = x * 10 + 7;
      }
      a.push_back(x);
    }
  }
  sort(a.begin(), a.end());
  cin >> vl >> vr;
  cin >> pl >> pr;
  int k;
  cin >> k;
  double res = 0;
  for (int i = 0; i + k - 1 < a.size(); ++i) {
    int la = 0;
    if (i > 0) la = a[i - 1] + 1;
    int na = 1000000001;
    if (i + k - 1 < a.size() - 1) na = a[i + k] - 1;
    if (k == 1 && a[i] >= vl && a[i] <= vr && a[i] >= pl && a[i] <= pr) {
      res -= (1.0 / (vr - vl + 1)) * (1.0 / (pr - pl + 1));
    }
    res += getp(la, na, a[i], a[i + k - 1]);
    swap(vl, pl);
    swap(vr, pr);
    res += getp(la, na, a[i], a[i + k - 1]);
  }
  printf("%.9lf", res);
  return 0;
}