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;
map<long long, long long> nex;
long long neext(long long x) {
  char lho[100];
  sprintf(lho, "%lld", x);
  string s = lho;
  reverse((s).begin(), (s).end());
  int i = 0;
  while (i < (int)s.length() && s[i] == '7') ++i;
  if (i == 0)
    s[i] = '7';
  else if (i < (int)s.length()) {
    s[i] = '7';
    for (int j = (int)0; j <= (int)i - 1; ++j) s[j] = '4';
  } else {
    int ln = s.length() + 1;
    s = "";
    for (int j = 0; j < (int)ln; ++j) s.push_back('4');
  }
  reverse((s).begin(), (s).end());
  long long ret = 0LL;
  for (int i = 0; i < (int)s.length(); ++i) {
    ret *= 10LL;
    ret += (long long)(s[i] - '0');
  }
  return ret;
}
int main() {
  scanf("%lld %lld %lld %lld %lld", &pl, &pr, &vl, &vr, &k);
  long long MAX = max(max(pl, pr), max(vl, vr));
  long long u1, uk, lho;
  long long bfor = 0LL, after = 0LL;
  u1 = uk = 4LL;
  for (int i = (int)2; i <= (int)k; ++i) {
    lho = neext(uk);
    nex[uk] = lho;
    uk = lho;
  }
  lho = neext(uk);
  nex[uk] = after = lho;
  nex[after] = neext(after);
  long long kount = 0LL;
  while (uk <= MAX) {
    long long p, v;
    if (u1 < pl)
      p = 0LL;
    else if (u1 <= pr)
      p = u1 - max(pl - 1LL, bfor);
    else {
      if (bfor >= pr)
        p = 0LL;
      else
        p = pr - max(bfor, pl - 1LL);
    }
    if (uk > vr)
      v = 0LL;
    else if (uk >= vl)
      v = min(vr + 1LL, after) - uk;
    else {
      if (after <= vl)
        v = 0LL;
      else
        v = min(vr + 1LL, after) - vl;
    }
    kount += p * v;
    if (u1 < vl)
      v = 0LL;
    else if (u1 <= vr)
      v = u1 - max(vl - 1LL, bfor);
    else {
      if (bfor >= vr)
        v = 0LL;
      else
        v = vr - max(bfor, vl - 1LL);
    }
    if (uk > pr)
      p = 0LL;
    else if (uk >= pl)
      p = min(pr + 1LL, after) - uk;
    else {
      if (after <= pl)
        p = 0LL;
      else
        p = min(pr + 1LL, after) - pl;
    }
    kount += p * v;
    if (k == 1 && u1 >= pl && u1 <= pr && u1 >= vl && u1 <= vr) --kount;
    bfor = u1;
    u1 = nex[u1];
    uk = after;
    after = nex[after];
    nex[after] = neext(after);
  }
  double ans =
      (double)kount / ((double)(pr - pl + 1LL) * (double)(vr - vl + 1LL));
  if (ans > 1) ans = 1;
  printf("%.20lf\n", ans);
  return 0;
}