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 p[1505], tot, p1, p2, v1, v2, k;
;
void dfs(long long x) {
  if (x > 1e9) return;
  p[tot++] = x;
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
int main() {
  scanf("%I64d%I64d%I64d%I64d%I64d", &p1, &p2, &v1, &v2, &k);
  tot = 0;
  dfs(0);
  sort(p, p + tot);
  p[tot] = 999999999;
  p[tot + 1] = 999999999;
  long long ans = 0;
  for (long long i = 1; i < tot; i++) {
    if (i + k - 1 > tot) break;
    long long x = p[i], y = p[i + k - 1];
    if (y > v2 && y > p2) break;
    long long l = 0, r = 0;
    l = max(l, min(p[k + i] - 1, v2) - max(v1, y) + 1);
    r = max(r, min(x, p2) - max(p1, p[i - 1] + 1) + 1);
    ans += l * r;
    l = r = 0;
    l = max(l, min(p[k + i] - 1, p2) - max(p1, y) + 1);
    r = max(r, min(x, v2) - max(v1, p[i - 1] + 1) + 1);
    ans += l * r;
    if (k == 1) {
      l = r = 0;
      l = max(l, min(p2, p[i]) - max(p1, p[i]) + 1);
      r = max(r, min(v2, p[i]) - max(v1, p[i]) + 1);
      ans -= l * r;
    }
  }
  double s = ans * 1.0 / ((p2 - p1 + 1) * (v2 - v1 + 1));
  printf("%.9lf\n", s);
  return 0;
}