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>
int sta[20];
int ln[2000], nl;
void toint(int now) {
  if (now == 0) return;
  int r = 0;
  for (int i = 1; i <= now; ++i) r = r * 10 + sta[i];
  ln[++nl] = r;
}
void gen(int now) {
  if (now == 10) return;
  toint(now);
  sta[now + 1] = 4;
  gen(now + 1);
  sta[now + 1] = 7;
  gen(now + 1);
  sta[now + 1] = 0;
}
int al, ar, bl, br, k;
int main(void) {
  gen(0);
  std::sort(ln + 1, ln + nl + 1);
  ln[++nl] = 2000000000;
  std::cin >> al >> ar >> bl >> br >> k;
  long long ansl = 0;
  double ans = 1;
  for (int le = 1, ri = k; ri < nl; ++le, ++ri) {
    int ln1 = ln[le], ln2 = ln[ri];
    long long a, b, c, d;
    a = b = c = d = -1;
    if (al <= ln1 && ln2 <= br) {
      a = std::min(ln1, ar) - std::max(al, ln[le - 1] + 1) + 1;
      b = std::min(br, ln[ri + 1] - 1) - std::max(ln2, bl) + 1;
      if (a < 0) a = 0;
      if (b < 0) b = 0;
      ansl += (a * b);
    }
    if (ln2 <= ar && bl <= ln1) {
      c = std::min(ln1, br) - std::max(bl, ln[le - 1] + 1) + 1;
      d = std::min(ar, ln[ri + 1] - 1) - std::max(ln2, al) + 1;
      if (c < 0) c = 0;
      if (d < 0) d = 0;
      ansl += (c * d);
    }
  }
  if (k == 1) {
    int le = std::max(al, bl), ri = std::min(ar, br);
    for (int i = 1; i < nl; ++i)
      if (le <= ln[i] && ln[i] <= ri) --ansl;
  }
  printf("%.10lf\n", ansl / ((double)(ar - al + 1) * (br - bl + 1)) + 1e-10);
  return 0;
}