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<int> v;
double prob(int s, int e, int l, int r) {
  int sz = e - s + 1, sect = max(min(e, r) - max(s, l) + 1, 0);
  return double(sect) / sz;
}
int main() {
  ios_base::sync_with_stdio(false);
  for (int i = 1; i <= 9; i++) {
    for (int j = 0; j < (1 << i); j++) {
      int x = 0;
      for (int k = i - 1; k >= 0; k--) x = x * 10 + (((j >> k) & 1) ? 7 : 4);
      v.push_back(x);
    }
  }
  int pl, pr, vl, vr, k;
  double ans = 0;
  cin >> pl >> pr >> vl >> vr >> k;
  for (int i = 0; i + k <= v.size(); i++) {
    if (v[i] > pr && v[i] > vr) break;
    int sl = (i ? v[i - 1] + 1 : 1), sr = v[i];
    int el = v[i + k - 1], er = (i + k < v.size() ? v[i + k] - 1 : 1e9);
    double now = 0;
    now += prob(pl, pr, sl, sr) * prob(vl, vr, el, er);
    now += prob(vl, vr, sl, sr) * prob(pl, pr, el, er);
    if (k == 1) now -= prob(pl, pr, v[i], v[i]) * prob(vl, vr, v[i], v[i]);
    ans += now;
  }
  cout << setprecision(10) << fixed << ans << endl;
  return 0;
}