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> Lucky;
long long pl, pr;
long long vl, vr;
long long k;
void max_self(long long& x, long long y) { x = max(x, y); }
void min_self(long long& x, long long y) { x = min(x, y); }
long long Getp(long long l, long long r) {
  max_self(l, pl);
  min_self(r, pr);
  return l <= r ? r - l + 1 : 0;
}
long long Getv(long long l, long long r) {
  max_self(l, vl);
  min_self(r, vr);
  return l <= r ? r - l + 1 : 0;
}
int main() {
  Lucky.push_back(0);
  for (long long Len = 1; Len < 11; Len++)
    for (long long Mask = 0; Mask < (1 << Len); Mask++) {
      long long Sum = 0;
      for (long long j = 0; j < Len; j++)
        if (Mask & (1 << j))
          Sum = 10 * Sum + 7;
        else
          Sum = 10 * Sum + 4;
      Lucky.push_back(Sum);
    }
  sort(Lucky.begin(), Lucky.end());
  cin >> pl >> pr >> vl >> vr >> k;
  long long Sum = 0;
  for (long long i = 1; i + k < Lucky.size(); i++) {
    Sum += Getp(Lucky[i - 1] + 1, Lucky[i]) *
               Getv(Lucky[i + k - 1], Lucky[i + k] - 1) +
           Getv(Lucky[i - 1] + 1, Lucky[i]) *
               Getp(Lucky[i + k - 1], Lucky[i + k] - 1);
    if (k == 1) Sum -= Getp(Lucky[i], Lucky[i]) * Getv(Lucky[i], Lucky[i]);
  }
  long double Ans =
      (long double)(Sum) / ((long double)((pr - pl + 1.0) * (vr - vl + 1.0)));
  printf("%.12f", (double)Ans);
}