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> g;
long long from, to;
void recurse(long long n) {
  if (n > to) return;
  if (n >= from && n <= to) {
    g.push_back(n);
  }
  recurse(n * 10 + 4);
  recurse(n * 10 + 7);
}
long long get(long long f, long long t, long long ff, long long tt) {
  if (ff > t || tt < f) return 0;
  ff = max(ff, f);
  tt = min(t, tt);
  return tt - ff + 1;
}
int main() {
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  from = min(pl, vl), to = max(pr, vr);
  recurse(4);
  recurse(7);
  sort(g.begin(), g.end());
  double res = 0.0;
  if ((int)g.size() < k) {
    puts("0.0");
    return 0;
  }
  long long total = 0;
  long long x, y, p, q, r, s;
  long long denom;
  denom = (pr - pl + 1) * (vr - vl + 1);
  int i, j, sz = g.size();
  i = 0;
  j = i + k - 1;
  while (1) {
    if (i == 0)
      p = from;
    else
      p = g[i - 1] + 1;
    q = g[i];
    r = g[j];
    if (j + 1 < (int)g.size())
      s = g[j + 1] - 1;
    else
      s = to;
    x = get(pl, pr, p, q);
    y = get(vl, vr, r, s);
    total = x * y;
    x = get(pl, pr, r, s);
    y = get(vl, vr, p, q);
    total += x * y;
    int chk = 0;
    if (g[i] >= max(pl, p)) chk++;
    if (g[i] <= max(pr, q)) chk++;
    if (g[i] >= max(vl, r)) chk++;
    if (g[i] <= max(vr, s)) chk++;
    if (g[i] >= max(pl, r)) chk++;
    if (g[i] <= max(pr, s)) chk++;
    if (g[i] >= max(vl, p)) chk++;
    if (g[i] <= max(vr, q)) chk++;
    if (chk == 8) {
      total--;
    }
    res += (total * 1.0 / denom);
    i++;
    j = i + k - 1;
    if (i >= sz || j >= sz) break;
  }
  printf("%.12lf\n", res);
  return 0;
}