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;
inline void performvll(vector<long long> &vll) {
  int i;
  for (i = 0; i < vll.size(); ++i) cout << vll[i] << " ";
  cout << endl;
}
inline void nextlucky(string &s, long long idx) {
  if (s[idx] >= '7') {
    s[idx] = '4';
    if (idx == 0)
      s.insert(s.begin(), '4');
    else
      nextlucky(s, idx - 1);
  } else if (s[idx] >= '4') {
    s[idx] = '7';
  } else {
    s[idx] = '4';
  }
}
inline long long nextlucky(long long lf) {
  if (lf <= 3)
    return 4;
  else {
    long long lastdgt, i;
    string s;
    while (lf != 0) {
      lastdgt = lf % 10;
      s.insert(s.begin(), (char)(lastdgt + '0'));
      lf /= 10;
    }
    nextlucky(s, s.size() - 1);
    long long result = 0, base = 1;
    for (i = s.size() - 1; i >= 0; --i) {
      result += (base * (int)(s[i] - '0'));
      base *= 10;
    }
    return result;
  }
}
inline long long lucky(long long pl, long long pr, long long vl, long long vr,
                       long long sleft, long long sright, long long lb,
                       long long rb) {
  long long result = 0;
  if (sleft < pl)
    result = 0;
  else if (sleft > pr) {
    if (lb > pr)
      result = 0;
    else {
      if (pl > lb)
        result = (pr - pl + 1);
      else
        result = (pr - lb);
    }
  } else {
    if (lb > pr)
      result = 0;
    else {
      if (pl > lb)
        result = (sleft - pl + 1);
      else
        result = (sleft - lb);
    }
  }
  if (sright > vr)
    result *= 0;
  else if (sright < vl) {
    if (rb < vl)
      result *= 0;
    else {
      if (rb > vr)
        result *= (vr - vl + 1);
      else
        result *= (rb - vl);
    }
  } else {
    if (rb < vl)
      result *= 0;
    else {
      if (rb > vr)
        result *= (vr - sright + 1);
      else
        result *= (rb - sright);
    }
  }
  return result;
}
int main() {
  long long pl, pr, vl, vr, k, minall, maxall, lf = 0, lbef = 0, i, adding, lb,
                                               rb;
  double all, now = 0;
  cin >> pl >> pr >> vl >> vr >> k;
  all = (double)(pr - pl + 1) * (double)(vr - vl + 1);
  minall = min(min(min(pl, pr), vl), vr);
  maxall = max(max(max(pl, pr), vl), vr);
  vector<long long> vlucky;
  while (lf < minall) {
    lbef = lf;
    lf = nextlucky(lf);
  }
  while (lf <= maxall) {
    vlucky.push_back(lf);
    lf = nextlucky(lf);
  }
  vector<long long> s = vlucky;
  for (i = 0; i <= s.size() - k; ++i) {
    if (i == 0)
      lb = -1;
    else
      lb = s[i - 1];
    if (i + k - 1 == s.size() - 1)
      rb = 90000000000;
    else
      rb = s[i + k];
    if (s[i] >= pl && s[i + k - 1] <= vr &&
        !(pl == pr && vl == vr && pl > vl)) {
      adding = lucky(pl, pr, vl, vr, s[i], s[i + k - 1], lb, rb);
      now += adding;
    }
    if (s[i] >= vl && s[i + k - 1] <= pr &&
        !(pl == pr && vl == vr && pl <= vl)) {
      adding = lucky(vl, vr, pl, pr, s[i], s[i + k - 1], lb, rb);
      now += adding;
    }
    if (k == 1 && (pl != pr) && (vl != vr) && (s[i] >= pl && s[i] <= pr) &&
        (s[i] >= vl) && (s[i] <= vr))
      now -= 1;
  }
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(17);
  cout << (double)(now / all) << endl;
  return 0;
}