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;
static const double EPS = 1e-8;
int main() {
  istream &fin = cin;
  ostream &fout = cout;
  FILE *fpin = stdin;
  FILE *fpout = stdout;
  vector<int> lucky;
  queue<int> q;
  q.push(0);
  while (!q.empty()) {
    int qt = q.front();
    q.pop();
    lucky.push_back(qt);
    if (qt * 10LL + 4 <= 1000000000) q.push(qt * 10 + 4);
    if (qt * 10LL + 7 <= 1000000000) q.push(qt * 10 + 7);
  }
  lucky.push_back(1000000001);
  int pl, pr, vl, vr, k;
  fin >> pl >> pr >> vl >> vr >> k;
  int n = lucky.size();
  double res = 0;
  for (int i = 0; i + k + 1 < n; i++) {
    int lrange = min(pr, lucky[i + 1]) - max(pl - 1, lucky[i]);
    int rrange = min(vr, lucky[i + k + 1] - 1) - max(vl - 1, lucky[i + k] - 1);
    lrange = max(lrange, 0);
    rrange = max(rrange, 0);
    res += ((double)lrange / (pr - pl + 1)) * ((double)rrange / (vr - vl + 1));
    lrange = min(vr, lucky[i + 1]) - max(vl - 1, lucky[i]);
    rrange = min(pr, lucky[i + k + 1] - 1) - max(pl - 1, lucky[i + k] - 1);
    lrange = max(lrange, 0);
    rrange = max(rrange, 0);
    res += ((double)lrange / (vr - vl + 1)) * ((double)rrange / (pr - pl + 1));
    if (k == 1) {
      if (pl <= lucky[i + 1] && lucky[i + 1] <= pr && vl <= lucky[i + 1] &&
          lucky[i + 1] <= vr)
        res -= (1. / (pr - pl + 1)) * (1. / (vr - vl + 1));
    }
  }
  fout << fixed << setprecision(9) << res << endl;
}