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;
long long ans, k, pl, pr, vl, vr, len, num[10000005];
void dfs(long long now) {
  if (now > max(pr, vr)) return;
  if (now >= min(pl, vl)) num[++len] = now;
  dfs(now * 10 + 4);
  dfs(now * 10 + 7);
}
signed main() {
  scanf("%I64d%I64d%I64d%I64d%I64d", &pl, &pr, &vl, &vr, &k);
  dfs(0);
  sort(num + 1, num + len + 1);
  num[len + 1] = 1e18;
  for (long long i = 1; i + k - 1 <= len; i++) {
    if (num[i] >= pl && num[i + k - 1] <= vr) {
      long long l1 = max(pl, num[i - 1] + 1), r1 = min(pr, num[i]);
      long long l2 = max(vl, num[i + k - 1]), r2 = min(num[i + k] - 1, vr);
      if (l1 <= r1 && l2 <= r2) ans += (r1 - l1 + 1) * (r2 - l2 + 1);
    }
    if (num[i] >= vl && num[i + k - 1] <= pr) {
      long long l1 = max(vl, num[i - 1] + 1), r1 = min(vr, num[i]);
      long long l2 = max(pl, num[i + k - 1]), r2 = min(num[i + k] - 1, pr);
      if (l1 <= r1 && l2 <= r2) ans += (r1 - l1 + 1) * (r2 - l2 + 1);
    }
  }
  for (long long i = 1; i <= len; i++)
    if (num[i] >= pl && num[i] <= pr && num[i] >= vl && num[i] <= vr && k == 1)
      ans--;
  long long all = (pr - pl + 1) * (vr - vl + 1);
  printf("%.12lf\n", (double)ans / (double)all);
  return 0;
}