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 int> L;
void pro(int n) {
  for (int i = 0; i < (1 << n); i++) {
    long long int p = 0;
    for (int j = 0; j < n; j++) {
      if ((i & (1 << j)) == 0)
        p = p * 10 + 4;
      else
        p = p * 10 + 7;
    }
    if (p <= 10000000000ll) L.push_back(p);
  }
}
long long int x;
long long int y;
long long int den = 0;
double un(long long int a, long long int b, long long int c, long long int d) {
  x = max(a, c);
  y = min(b, d);
  if (x > y)
    return 0;
  else
    return y - x + 1;
}
int is(long long int a) {
  for (int i = 0; i < L.size(); i++)
    if (L[i] == a) return i;
  return -1;
}
int main() {
  L.push_back(0);
  for (int i = 1; i < 11; i++) pro(i);
  sort(L.begin(), L.end());
  L.resize(unique(L.begin(), L.end()) - L.begin());
  long long int pl, pr, vl, vr, k;
  L.push_back(1000000000);
  cin >> pl >> pr >> vl >> vr >> k;
  long long int res = 0;
  for (int i = 1; i <= L.size() - k; i++) {
    res +=
        un(L[i - 1] + 1, L[i], pl, pr) * un(L[i + k - 1], L[i + k] - 1, vl, vr);
    res +=
        un(L[i - 1] + 1, L[i], vl, vr) * un(L[i + k - 1], L[i + k] - 1, pl, pr);
    if (k == 1 && L[i] <= pr && L[i] <= vr && L[i] >= pl && L[i] >= vl) res--;
  }
  cout << setiosflags(ios::showpoint | ios::fixed) << setprecision(10)
       << 1.0 * res / (vr - vl + 1) / (pr - pl + 1) << endl;
  return 0;
}