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;
void cal(long long l, long long r, long long* array1, long long* array2) {
  long long pre = 0LL;
  long long cnt = 0LL;
  memset(array1, 0LL, sizeof(array1));
  memset(array2, 0LL, sizeof(array2));
  for (int bit = 1; bit <= 10; ++bit) {
    for (int i = 0; i < (1 << bit); ++i) {
      long long cur = 0LL;
      for (int j = bit - 1; j >= 0; --j) {
        if ((1 << j) & i) {
          cur = cur * 10LL + 7LL;
        } else {
          cur = cur * 10LL + 4LL;
        }
      }
      if (cur < l) {
        pre = cur;
        cnt++;
      } else {
        if (cur > r && pre < l) {
          array1[cnt] = r - l + 1;
          array2[cnt] = r - l + 1;
        } else {
          long long tl, tr;
          tl = max(pre, l);
          tr = min(cur - 1, r);
          if (tl <= tr) {
            array1[cnt] = tr - tl + 1;
          }
          tl = max(pre + 1, l);
          tr = min(cur, r);
          if (tl <= tr) {
            array2[cnt] = tr - tl + 1;
          }
        }
        pre = cur;
        cnt++;
      }
    }
  }
}
int main() {
  long long pl, pr, vl, vr, k;
  long long p1[2000], p2[2000], v1[2000], v2[2000];
  while (cin >> pl >> pr >> vl >> vr >> k) {
    cal(pl, pr, p1, p2);
    cal(vl, vr, v1, v2);
    long long ans = 0LL;
    for (int num = 1024; num >= k; --num) {
      ans += v1[num] * p2[num - k];
      ans += p1[num] * v2[num - k];
      if (k == 1) {
        if (v1[num] * p2[num - k] > 0 && p1[num] * v2[num - k] > 0) {
          ans--;
        }
      }
    }
    double total = (double)((pr - pl + 1) * (vr - vl + 1));
    printf("%.10lf\n", (double)ans / total);
  }
  return 0;
}