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<int> lk;
void make(int num, int len) {
  if (len == 10) return;
  if (num != 0) lk.push_back(num);
  make(num * 10 + 4, len + 1);
  make(num * 10 + 7, len + 1);
}
double gao(int l1, int l2, int r1, int r2) {
  double l = max(l1, r1), r = min(l2, r2);
  return max(0.0, r - l + 1);
}
int main() {
  make(0, 0);
  sort(lk.begin(), lk.end());
  int p1, p2, v1, v2, k;
  while (cin >> p1 >> p2 >> v1 >> v2 >> k) {
    int c1 = max(p1, v1), c2 = min(c2, v2);
    double tot = (p2 - p1 + 1.0) * (v2 - v1 + 1.0), sum = 0;
    int sz = lk.size();
    for (int i = 0; i + k - 1 < sz; i++) {
      int l1, l2 = lk[i], r1 = lk[i + k - 1], r2;
      if (i == 0)
        l1 = 0;
      else
        l1 = lk[i - 1] + 1;
      if (i + k == sz)
        r2 = 1000000000;
      else
        r2 = lk[i + k] - 1;
      sum += gao(l1, l2, p1, p2) * gao(r1, r2, v1, v2);
      sum += gao(l1, l2, v1, v2) * gao(r1, r2, p1, p2);
      sum -= gao(c1, c2, max(l1, r1), min(l2, r2));
    }
    printf("%.12f\n", sum / tot);
  }
  return 0;
}