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;
vector<long long> a;
long long pl, pr, vl, vr, k;
void gen(long long cur) {
  a.push_back(cur);
  if (cur * 10LL + 4LL <= 2000000000LL) gen(cur * 10LL + 4LL);
  if (cur * 10LL + 7LL <= 2000000000LL) gen(cur * 10LL + 7LL);
}
double prob(int l, int r, int cl, int cr) {
  if (cl > r) return 0;
  if (cr < l) return 0;
  int tl = max(l, cl);
  int tr = min(r, cr);
  return (tr - tl + 1.0) / (r - l + 1.0);
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  gen(4);
  gen(7);
  sort(a.begin(), a.end());
  double res = 0;
  for (int i = 0; i + k - 1 < a.size(); i++) {
    int st = a[i];
    int fi = a[i + k - 1];
    if (fi > 1000000000) break;
    int dn = 1;
    if (i != 0) dn = a[i - 1] + 1;
    int up = 1000000000;
    if (i + k < a.size()) up = a[i + k] - 1;
    res += prob(pl, pr, dn, st) * prob(vl, vr, fi, up);
    res += prob(vl, vr, dn, st) * prob(pl, pr, fi, up);
    if (k == 1) {
      if (st >= pl && st <= pr && st >= vl && st <= vr)
        res -= 1 / (double)(pr - pl + 1) / (double)(vr - vl + 1);
    }
  }
  printf("%.10lf\n", res);
  return 0;
}