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 mm[2000];
int tt = 0;
void dfs(long long k) {
  if (k > 1000000000) return;
  mm[tt++] = k;
  dfs(k * 10 + 4);
  dfs(k * 10 + 7);
}
double jiao(long long s1, long long e1, long long s2, long long e2) {
  if (s1 < s2) s1 = s2;
  if (e1 > e2) e1 = e2;
  if (s1 > e1) return 0;
  return e1 - s1 + 1;
}
int main() {
  int a, b, i, j, k;
  long long pl, pr, vl, vr;
  double sum, ok, s1, s2;
  dfs(0);
  mm[tt++] = 99999999999;
  sort(mm, mm + tt);
  while (scanf("%I64d %I64d %I64d %I64d %d", &pl, &pr, &vl, &vr, &k) != EOF) {
    ok = 0;
    sum = (pr - pl + 1) * (vr - vl + 1);
    for (i = 1; i + k - 1 < tt; ++i) {
      if (k == 1 && mm[i] <= min(pr, vr) && mm[i] >= max(pl, vl)) ok--;
      j = i + k - 1;
      ok += jiao(mm[i - 1] + 1, mm[i], pl, pr) *
            jiao(mm[j], mm[j + 1] - 1, vl, vr);
      ok += jiao(mm[i - 1] + 1, mm[i], vl, vr) *
            jiao(mm[j], mm[j + 1] - 1, pl, pr);
    }
    printf("%.15lf\n", ok / sum);
  }
  return 0;
}