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, ans;
vector<long long> lucky;
vector<long long>::iterator it, it2;
void dfs(long long n) {
  if (n > (1e10)) return;
  lucky.push_back(n);
  dfs(10 * n + 4);
  dfs(10 * n + 7);
}
void doit(long long pl, long long pr, long long vl, long long vr, long long k) {
  long long p = pl, v = vl;
  bool fp = false, fv = false;
  while (p <= pr && v <= vr) {
    if (!fp) it = lower_bound(lucky.begin(), lucky.end(), p);
    if (!fv) it2 = upper_bound(lucky.begin(), lucky.end(), v);
    long long now = it2 - it;
    if (now == k) {
      ans += ((min(pr, *it) - max(pl - 1, *(it - 1))) *
              (min(vr, (*it2) - 1) - max(vl - 1, *(it2 - 1) - 1)));
      v = *it2;
      it2++;
      fv = true;
      if (!fp) {
        it = upper_bound(lucky.begin(), lucky.end(), p) + 1;
        p = *it;
        fp = true;
      } else {
        it++;
        p = *it;
      }
      continue;
    }
    if (now < k) {
      v = *it2;
      it2++;
      fv = true;
      continue;
    }
    if (now > k) {
      if (!fp) {
        it = upper_bound(lucky.begin(), lucky.end(), p) + 1;
        p = *it;
        fp = true;
      } else {
        it++;
        p = *it;
      }
      continue;
    }
  }
}
long long esht(long long l1, long long r1, long long l2, long long r2) {
  if (l2 > r1 || l1 > r2) return 0;
  return -max(l1, l2) + min(r1, r2) + 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> pl >> pr >> vl >> vr >> k;
  dfs(0ll);
  sort(lucky.begin(), lucky.end());
  int i = 1;
  while (i + k < lucky.size()) {
    bool f = false;
    if (k == 1)
      ans += esht(lucky[i - 1] + 1, lucky[i], pl, pr) *
             esht(lucky[i], lucky[i + 1] - 1, vl, vr);
    if (k == 1)
      ans += esht(lucky[i - 1] + 1, lucky[i], vl, vr) *
             esht(lucky[i], lucky[i + 1] - 1, pl, pr);
    if (k == 1)
      ans -=
          esht(lucky[i], lucky[i], vl, vr) * esht(lucky[i], lucky[i], pl, pr);
    if (k != 1)
      ans += esht(lucky[i - 1] + 1, lucky[i], pl, pr) *
             esht(lucky[i + k - 1], lucky[i + k] - 1, vl, vr);
    if (k != 1)
      ans += esht(lucky[i - 1] + 1, lucky[i], vl, vr) *
             esht(lucky[i + k - 1], lucky[i + k] - 1, pl, pr);
    i++;
  }
  cout << setprecision(13) << fixed
       << (double)ans / (vr - vl + 1) / (pr - pl + 1);
}