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;
long long pl, pr, vl, vr, k;
int l[100000], lsz;
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  long long pl1 = pl, vl1 = vl;
  for (int len = 1; len <= 9; len++) {
    for (int p = 0; p < (1 << len); ++p) {
      int x = 0;
      for (int i = 0; i < len; ++i)
        if (p & (1 << i))
          x = x * 10 + 7;
        else
          x = x * 10 + 4;
      l[lsz++] = x;
    }
  }
  sort(l, lsz + l);
  long long ans = 0;
  for (int i = 0; i + k - 1 < lsz && l[i] <= max(vr, pr); ++i) {
    long long left = l[i], right = l[i + k - 1], lf, rh;
    if (i == 0)
      lf = 0;
    else
      lf = l[i - 1] + 1;
    if (i + k < lsz)
      rh = l[i + k] - 1;
    else
      rh = (1 << 30);
    if (!(min(left, pr) - max(pl, lf) + 1 <= 0 ||
          min(vr, rh) - max(right, vl1) + 1 <= 0)) {
      ans += (min(left, pr) - max(pl, lf) + 1) *
             (min(vr, rh) - max(right, vl1) + 1);
    }
    if (!(min(left, vr) - max(vl, lf) + 1 <= 0 ||
          min(rh, pr) - max(right, pl1) + 1 <= 0)) {
      ans += (min(left, vr) - max(vl, lf) + 1) *
             (min(rh, pr) - max(right, pl1) + 1);
    }
    pl = max(left + 1, pl);
    vl = max(left + 1, vl);
    if (pl > pr || vl > vr) break;
  }
  if (k == 1)
    for (int i = 0; i < lsz; ++i)
      if (pl1 <= l[i] && l[i] <= pr && vl1 <= l[i] && l[i] <= vr) ans--;
  printf("%.20lf",
         (double)ans / ((vr * 1.0 - vl1 + 1.0) * (pr * 1.0 - pl1 + 1.0)));
  return 0;
}