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 pl, pr, vl, vr, k, a[10020], t, x, y, z = 0;
int c = 0;
void dfs(long long x) {
  if (x > 1e10) return;
  a[c++] = x;
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
long long fun(long long pl, long long pr, long long vl, long long vr) {
  return max(min(pr, vr) - max(pl, vl) + 1, 0LL);
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  t = (pr - pl + 1) * (vr - vl + 1);
  dfs(0);
  sort(a, a + c);
  for (int i = 1, j; i + k < c; i++) {
    j = i + k;
    x = fun(a[i - 1] + 1, a[i], pl, pr);
    y = fun(a[j - 1], a[j] - 1, vl, vr);
    z += x * y;
    x = fun(a[i - 1] + 1, a[i], vl, vr);
    y = fun(a[j - 1], a[j] - 1, pl, pr);
    z += x * y;
    if (k == 1 && pl <= a[i] && a[i] <= pr && vl <= a[i] && a[i] <= vr) z--;
  }
  printf("%.15lf\n", 1.0 * z / t);
  return 0;
}