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;
vector<long long> vec;
void dfs(long long x) {
  vec.push_back(x);
  if ((x * 10 + 4) <= 1e9) dfs(x * 10 + 4);
  if ((x * 10 + 7) <= 1e9) dfs(x * 10 + 7);
  return;
}
int main() {
  dfs(4);
  dfs(7);
  sort(vec.begin(), vec.end());
  long long pl, pr, vl, vr, k;
  scanf("%lld %lld %lld %lld %lld", &pl, &pr, &vl, &vr, &k);
  double res = 0.0;
  int sz = vec.size();
  for (int i = 0; i + k - 1 < sz; i++) {
    if (pl > vec[i]) continue;
    if (vr < vec[i + k - 1]) continue;
    long long l, r;
    if (i)
      l = min(vec[i], pr) - max(vec[i - 1] + 1, pl) + 1;
    else
      l = min(vec[i], pr) - pl + 1;
    l = max(l, 0ll);
    if ((i + k - 1) == sz - 1)
      r = vr - max(vec[i + k - 1], vl) + 1;
    else
      r = min(vr, vec[i + k] - 1) - max(vec[i + k - 1], vl) + 1;
    r = max(r, 0ll);
    res += l * r;
  }
  swap(pl, vl);
  swap(pr, vr);
  for (int i = 0; i + k - 1 < sz; i++) {
    if (pl > vec[i]) continue;
    if (vr < vec[i + k - 1]) continue;
    long long l, r;
    if (i)
      l = min(vec[i], pr) - max(vec[i - 1] + 1, pl) + 1;
    else
      l = min(vec[i], pr) - pl + 1;
    l = max(l, 0ll);
    if ((i + k - 1) == sz - 1)
      r = vr - max(vec[i + k - 1], vl) + 1;
    else
      r = min(vr, vec[i + k] - 1) - max(vec[i + k - 1], vl) + 1;
    r = max(r, 0ll);
    res += l * r;
    if (k == 1) res -= 1.0;
  }
  double deno = (pr - pl + 1) * 1.0 * (vr - vl + 1);
  res /= deno;
  printf("%.10f\n", res);
  return 0;
}