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 pl, pr, vl, vr, n, m, c[100001], cnt;
int main() {
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &m);
  cnt = 0;
  c[0] = 0;
  for (int i = 0; i <= cnt; i++) {
    if (c[i] * 10LL + 4 <= 1000000000) c[++cnt] = c[i] * 10 + 4;
    if (c[i] * 10LL + 7 <= 1000000000) c[++cnt] = c[i] * 10 + 7;
  }
  c[++cnt] = 1000000001;
  double ans = 0;
  for (int i = 1; i + m - 1 < cnt; i++)
    for (int j = 1; j <= 2; j++) {
      int L1 = max(pl, c[i - 1] + 1), R1 = min(pr, c[i]),
          L2 = max(vl, c[i + m - 1]), R2 = min(vr, c[i + m] - 1);
      if (L1 <= R1 && L2 <= R2) {
        if (m == 1 && j == 2)
          ans += 1.0 * (R1 - L1 + 1) / (pr - pl + 1) * (R2 - L2 + 1) /
                     (vr - vl + 1) -
                 1.0 / (pr - pl + 1) / (vr - vl + 1);
        else
          ans += 1.0 * (R1 - L1 + 1) / (pr - pl + 1) * (R2 - L2 + 1) /
                 (vr - vl + 1);
      }
      swap(pl, vl);
      swap(pr, vr);
    }
  printf("%.10f\n", ans);
}