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> luck;
void back(long long num, int pos) {
  if (pos == 10) {
    return;
  }
  luck.push_back(num * 10LL + 4LL);
  back(num * 10LL + 4LL, pos + 1);
  luck.push_back(num * 10LL + 7LL);
  back(num * 10LL + 7LL, pos + 1);
}
long long calc(long long l1, long long r1, long long l2, long long r2) {
  if (r1 < l2) return 0;
  if (r2 < l1) return 0;
  if (l1 < l2) {
    return (min(r1, r2) + 1 - l2);
  } else {
    return (min(r1, r2) + 1 - l1);
  }
}
int main() {
  back(0, 0);
  sort(luck.begin(), luck.end());
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  long long tot = (pr - pl + 1) * (vr - vl + 1);
  long long way = 0;
  for (int i = 0; i < luck.size() - 1; i++) {
    int j = i + k - 1;
    if (j + 1 >= luck.size()) break;
    if (i != 0) {
      long long p1 = calc(luck[i - 1] + 1, luck[i], pl, pr);
      long long p2 = calc(luck[j], luck[j + 1] - 1, vl, vr);
      way = way + p1 * p2;
      p1 = calc(luck[i - 1] + 1, luck[i], vl, vr);
      p2 = calc(luck[j], luck[j + 1] - 1, pl, pr);
      way = way + p1 * p2;
      if (k == 1 && luck[i] >= vl && luck[i] <= vr && luck[i] >= pl &&
          luck[i] <= pr)
        way--;
    } else {
      long long p1 = calc(1, luck[i], pl, pr);
      long long p2 = calc(luck[j], luck[j + 1] - 1, vl, vr);
      way = way + p1 * p2;
      p1 = calc(1, luck[i], vl, vr);
      p2 = calc(luck[j], luck[j + 1] - 1, pl, pr);
      way = way + p1 * p2;
      if (k == 1 && luck[i] >= vl && luck[i] <= vr && luck[i] >= pl &&
          luck[i] <= pr)
        way--;
    }
  }
  double a1 = way;
  double b1 = tot;
  a1 = a1 / b1;
  printf("%.10lf\n", a1);
  return 0;
}