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;
int tens[] = {1,      10,      100,      1000,      10000,
              100000, 1000000, 10000000, 100000000, 1000000000};
vector<int> nums;
void getc(int b, int start) {
  int c = 0;
  int ret = 0;
  int t = b;
  while (t) {
    t /= 10;
    c++;
  }
  int base = 0;
  int curr = 0;
  if (c == 10) --c;
  for (int digCount = 1; digCount <= c; ++digCount) {
    base = 0;
    for (int k = 0; k < digCount; ++k) base += 4 * tens[k];
    for (int i = 0; i < 1 << (digCount); ++i) {
      curr = base;
      for (int j = 0; j < digCount; ++j) {
        if (1 << j & i) curr += tens[j] * 3;
      }
      if (curr <= b) {
        if (curr >= start) nums.push_back(curr);
      } else
        return;
    }
  }
  return;
}
unsigned long long inter(unsigned long long l1, unsigned long long r1,
                         unsigned long long l2, unsigned long long r2) {
  unsigned long long l = l1 > l2 ? l1 : l2;
  unsigned long long r = r1 < r2 ? r1 : r2;
  return r >= l > 0 ? r - l + 1 : 0;
}
int main() {
  int vl, vr, pl, pr, k, r, l;
  scanf("%d%d%d%d%d", &vl, &vr, &pl, &pr, &k);
  r = max(vr, pr);
  l = min(vl, pl);
  getc(r, l);
  unsigned long long omega = (unsigned long long)(vr - vl + 1) * (pr - pl + 1);
  if (nums.size() < k) {
    puts("0");
    return 0;
  }
  unsigned long long c = 0;
  k--;
  unsigned long long l1, l2, r1, r2;
  for (int i = 0; i + k < nums.size(); ++i) {
    l1 = i > 0 ? nums[i - 1] + 1 : 0;
    r1 = nums[i];
    l2 = nums[i + k];
    r2 = i + k + 1 < nums.size() ? nums[i + k + 1] - 1 : 1000000001;
    c += inter(l1, r1, vl, vr) * inter(l2, r2, pl, pr);
    c += inter(l1, r1, pl, pr) * inter(l2, r2, vl, vr);
    if (k == 0 && (vl <= r1 && vr >= r1) && (pl <= r1 && pr >= r1)) c--;
  }
  printf("%.12lf\n", (double)c / omega);
  return 0;
}