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;
const int INF = 2000000000;
const double PI = 3.141592653589793;
const double EPS = 1e-9;
vector<long long> mas;
long long pl, pr, vl, vr, k;
void Generate() {
  mas.push_back(4);
  mas.push_back(7);
  int p = 0;
  int q = 10;
  while (q < 1000000000) {
    int i = p;
    p = mas.size();
    for (; i < p; ++i) {
      mas.push_back(4 * q + mas[i]);
      mas.push_back(7 * q + mas[i]);
    }
    q *= 10;
  }
}
bool IsIn(long long x, long long l, long long r) { return x >= l && x <= r; }
bool F(long long l1, long long r1, long long l2, long long r2) {
  return l1 >= l2 && r1 <= r2 || l2 >= l1 && r2 >= r1;
}
int main() {
  Generate();
  mas.push_back(0);
  mas.push_back(1000000001);
  sort(mas.begin(), mas.end());
  cin >> pl >> pr >> vl >> vr >> k;
  double prob = 0;
  for (int i = 1, j = k; j < mas.size() - 1; ++i, ++j) {
    long long _pl = max(pl, mas[i - 1] + 1);
    long long _pr = min(pr, mas[i]);
    long long _vl = max(vl, mas[j]);
    long long _vr = min(vr, mas[j + 1] - 1);
    long long _pCnt = 0;
    if (IsIn(_pl, mas[i - 1] + 1, mas[i]) &&
        IsIn(_pr, mas[i - 1] + 1, mas[i]) && _pr >= _pl)
      _pCnt = _pr - _pl + 1;
    long long _vCnt = 0;
    if (IsIn(_vl, mas[j], mas[j + 1] - 1) &&
        IsIn(_vr, mas[j], mas[j + 1] - 1) && _vr >= _vr)
      _vCnt = _vr - _vl + 1;
    prob += (_pCnt / (double)(pr - pl + 1)) * (_vCnt / (double)(vr - vl + 1));
    long long __pl = max(vl, mas[i - 1] + 1);
    long long __pr = min(vr, mas[i]);
    long long __vl = max(pl, mas[j]);
    long long __vr = min(pr, mas[j + 1] - 1);
    long long __pCnt = 0;
    if (IsIn(__pl, mas[i - 1] + 1, mas[i]) &&
        IsIn(__pr, mas[i - 1] + 1, mas[i]) && __pr >= __pl)
      __pCnt = __pr - __pl + 1;
    long long __vCnt = 0;
    if (IsIn(__vl, mas[j], mas[j + 1] - 1) &&
        IsIn(__vr, mas[j], mas[j + 1] - 1) && __vr >= __vr)
      __vCnt = __vr - __vl + 1;
    prob += (__pCnt / (double)(vr - vl + 1)) * (__vCnt / (double)(pr - pl + 1));
    if (k == 1 && _pCnt && _vCnt && __pCnt && __vCnt && F(_pl, _pr, _vl, _vr) &&
        F(__pl, __pr, __vl, __vr))
      prob -= (1 / (double)(vr - vl + 1)) * (1 / (double)(pr - pl + 1));
  }
  printf("%.12f\n", min(prob, 1.0));
  return 0;
}