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> lucks;
long long pl, pr, vl, vr;
map<long long, bool> islucky;
int Lnumless[10000];
int Rnumless[10000];
int ILL[10000];
int ILR[10000];
double probvalL[10000];
double probvalR[10000];
int main() {
  for (int d = 1; d <= 10; d++) {
    for (int m = 0; m < (1 << d); m++) {
      long long number = 0;
      for (int j = 0; j < d; j++) {
        if ((m >> j) & 1)
          number = number * 10 + 7;
        else
          number = number * 10 + 4;
      }
      lucks.push_back(number);
      islucky[number] = true;
    }
  }
  sort(lucks.begin(), lucks.end());
  cin >> pl >> pr;
  cin >> vl >> vr;
  vector<long long> Left, Right;
  for (int j = 0; j < lucks.size(); j++) {
    if (lucks[j] >= pl && lucks[j] <= pr) Left.push_back(lucks[j]);
    if (lucks[j] >= vl && lucks[j] <= vr) Right.push_back(lucks[j]);
  }
  vector<pair<long long, long long> > Lint;
  int prev = pl;
  for (int j = 0; j < Left.size(); j++) {
    if (Left[j] > prev) Lint.push_back(make_pair(prev, Left[j] - prev));
    Lint.push_back(make_pair(Left[j], 1));
    prev = Left[j] + 1;
  }
  if (prev <= pr) Lint.push_back(make_pair(prev, pr - prev + 1));
  vector<pair<long long, long long> > Rint;
  prev = vl;
  for (int j = 0; j < Right.size(); j++) {
    if (Right[j] > prev) Rint.push_back(make_pair(prev, Right[j] - prev));
    Rint.push_back(make_pair(Right[j], 1));
    prev = Right[j] + 1;
  }
  if (prev <= vr) Rint.push_back(make_pair(prev, vr - prev + 1));
  for (int j = 0; j < Lint.size(); j++) {
    Lnumless[j] = 0;
    for (int i = 0; i < lucks.size(); i++) {
      if (lucks[i] <= Lint[j].first)
        Lnumless[j]++;
      else
        break;
    }
    if (islucky[Lint[j].first])
      ILL[j] = 1;
    else
      ILL[j] = 0;
    probvalL[j] = Lint[j].second / (pr - pl + 1.0);
  }
  for (int j = 0; j < Rint.size(); j++) {
    Rnumless[j] = 0;
    for (int i = 0; i < lucks.size(); i++) {
      if (lucks[i] <= Rint[j].first)
        Rnumless[j]++;
      else
        break;
    }
    if (islucky[Rint[j].first])
      ILR[j] = 1;
    else
      ILR[j] = 0;
    probvalR[j] = Rint[j].second / (vr - vl + 1.0);
  }
  double prob = 0;
  int reqd;
  cin >> reqd;
  for (int j = 0; j < Lint.size(); j++)
    for (int k = 0; k < Rint.size(); k++) {
      int numlucks;
      if (Lint[j].first > Rint[k].first)
        numlucks = Lnumless[j] - Rnumless[k] + ILR[k];
      if (Lint[j].first < Rint[k].first)
        numlucks = Rnumless[k] - Lnumless[j] + ILL[j];
      if (Lint[j].first == Rint[k].first) numlucks = ILL[j];
      if (numlucks == reqd) prob += probvalL[j] * probvalR[k];
    }
  printf("%1.10f\n", prob);
}