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 n, pl, pr, vl, vr;
vector<int> mklucky() {
  vector<int> u;
  u.push_back(0);
  for (int i = 0; i < 9; i++) {
    int m = 1 << i;
    for (int k = 0; k < m; k++) {
      u.push_back(u[m - 1 + k] * 10 + 4);
      u.push_back(u[m - 1 + k] * 10 + 7);
    }
  }
  return u;
}
double pin(int l, int r) {
  return max(min(pr, r) - max(l, pl) + 1, 0) * 1.0 / (pr - pl + 1);
}
double vin(int l, int r) {
  return max(min(vr, r) - max(l, vl) + 1, 0) * 1.0 / (vr - vl + 1);
}
double gao(int al, int ar, int bl, int br) {
  if (ar == bl)
    return pin(al, ar) * vin(bl, br) + pin(bl, br) * vin(al, ar) -
           pin(ar, ar) * vin(ar, ar);
  else
    return pin(al, ar) * vin(bl, br) + pin(bl, br) * vin(al, ar);
}
int main() {
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &n);
  vector<int> u = mklucky();
  u.push_back(1000000001);
  double ans = 0;
  for (size_t i = n; i + 1 < u.size(); i++) {
    ans += gao(u[i - n] + 1, u[i - n + 1], u[i], u[i + 1] - 1);
  }
  printf("%.14f\n", ans);
}