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>
int a[20000];
int pl, pr, vl, vr;
long long cal(int a1, int a2, int b1, int b2) {
  if (pl > a1) a1 = pl;
  if (pr < a2) a2 = pr;
  if (a1 > a2) return 0;
  if (vl > b1) b1 = vl;
  if (vr < b2) b2 = vr;
  if (b1 > b2) return 0;
  return (long long)(a2 - a1 + 1) * (b2 - b1 + 1);
}
int main() {
  int i, j, t, k;
  long long t1, t2;
  t = 0;
  a[0] = 0;
  t++;
  for (i = 0; i < t; i++) {
    if (a[i] >= 100000000) continue;
    a[t++] = a[i] * 10 + 4;
    a[t++] = a[i] * 10 + 7;
  }
  a[t] = 1000000001;
  scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &k);
  t2 = (long long)(pr - pl + 1) * (vr - vl + 1);
  t1 = 0;
  for (i = 1; i <= t - k; i++) {
    t1 += cal(a[i - 1] + 1, a[i], a[i + k - 1], a[i + k] - 1);
    t1 += cal(a[i + k - 1], a[i + k] - 1, a[i - 1] + 1, a[i]);
    if (k == 1) t1 -= cal(a[i], a[i], a[i], a[i]);
  }
  printf("%.10lf\n", (double)t1 / t2);
  return 0;
}