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> save;
void ini() {
  set<long long> s1;
  set<long long>::iterator it;
  for (int tam = 1; tam <= 11; tam++) {
    for (int i = 0; i < (1 << tam); i++) {
      long long num = 0;
      for (int j = 0; j < tam; j++) {
        num = num * 10 + ((i & (1 << j)) ? 7 : 4);
      }
      s1.insert(num);
    }
  }
  save.push_back(0);
  for (it = s1.begin(); it != s1.end(); it++) {
    save.push_back(*it);
  }
}
void imp(long long iniA, long long finA, long long iniB, long long finB) {
  cout << "[" << iniA << ":" << finA << "] --- [" << iniB << ":" << finB << "]"
       << endl;
}
long long iniC, finC;
long long busco(long long iniA, long long finA, long long iniB,
                long long finB) {
  if (finB < iniA) return 0;
  if (finA < iniB) return 0;
  finC = min(finA, finB);
  iniC = max(iniA, iniB);
  return min(finA, finB) - max(iniA, iniB) + 1;
}
int main() {
  ini();
  long long p_ini, p_fin, v_ini, v_fin, k;
  cin >> p_ini >> p_fin >> v_ini >> v_fin >> k;
  long long ans = 0;
  for (int i = 1; i + k < save.size() - 4; i++) {
    long long ini_x = save[i - 1] + 1, fin_x = save[i], ini_y = save[i + k - 1],
              fin_y = save[i + k] - 1;
    long long iniP, finP, iniS, finS;
    long long tt1, tt2, tt3 = 0, tt4 = 0;
    long long iniX, finX, iniY, finY;
    tt1 = busco(ini_x, fin_x, p_ini, p_fin);
    if (tt1 != 0) iniX = iniC, finX = finC;
    tt2 = busco(ini_y, fin_y, v_ini, v_fin);
    if (tt2 != 0) iniY = iniC, finY = finC;
    ans += tt1 * tt2;
    if (tt1 != 0 && tt2 != 0) {
      tt3 = busco(iniX, finX, iniY, finY);
      if (tt3 != 0) iniP = iniC, finP = finC;
    }
    tt1 = busco(ini_x, fin_x, v_ini, v_fin);
    if (tt1 != 0) iniX = iniC, finX = finC;
    tt2 = busco(ini_y, fin_y, p_ini, p_fin);
    if (tt2 != 0) iniY = iniC, finY = finC;
    ans += tt1 * tt2;
    if (tt1 != 0 && tt2 != 0) {
      tt4 = busco(iniX, finX, iniY, finY);
      if (tt4 != 0) iniS = iniC, finS = finC;
    }
    if (tt3 != 0 && tt4 != 0) {
      tt1 = busco(iniP, finP, iniS, finS);
      ans -= tt1;
    }
  }
  double tt = ans * 1.0 / ((p_fin - p_ini + 1) * (v_fin - v_ini + 1));
  printf("%.10lf\n", min(1.0, tt));
  return 0;
}