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;
void rec(long long n) {
  if (n > 1e9 + 10) return;
  v.push_back(n);
  rec(n * 10 + 4);
  rec(n * 10 + 7);
}
int main() {
  v.clear();
  rec(0);
  v.push_back(1e9 + 100);
  sort((v).begin(), (v).end());
  long long a1, a2, b1, b2;
  int k;
  cin >> a1 >> a2 >> b1 >> b2 >> k;
  long long tot = (a2 - a1 + 1) * (b2 - b1 + 1);
  long long sum = 0;
  for (int(i) = (1); (i) < (v.size()); (i)++) {
    if (i + k >= v.size()) break;
    int start = i;
    int end = i + k - 1;
    long long p1 = v[start - 1] + 1;
    long long p2 = v[start];
    long long q1 = v[end];
    long long q2 = v[end + 1] - 1;
    long long c1 = max(a1, p1);
    long long c2 = min(a2, p2);
    long long first = 0;
    if (c2 >= c1) first = c2 - c1 + 1;
    long long d1 = max(b1, q1);
    long long d2 = min(b2, q2);
    long long sec = 0;
    if (d2 >= d1) sec = d2 - d1 + 1;
    sum += sec * first;
    bool can = false;
    long long my1, my2;
    if (first != 0 && sec != 0) {
      can = true;
      my1 = max(c1, d1);
      my2 = min(c2, d2);
    }
    c1 = max(a1, q1);
    c2 = min(a2, q2);
    first = 0;
    if (c2 >= c1) first = c2 - c1 + 1;
    d1 = max(b1, p1);
    d2 = min(b2, p2);
    sec = 0;
    if (d2 >= d1) sec = d2 - d1 + 1;
    sum += sec * first;
    if (first != 0 && sec != 0 && can && my2 >= my1) {
      long long my3 = max(c1, d1);
      long long my4 = min(c2, d2);
      long long fin1 = max(my1, my3);
      long long fin2 = min(my2, my4);
      if (fin2 >= fin1) sum -= fin2 - fin1 + 1;
    }
  }
  cout << fixed << setprecision(10) << (long double)sum / tot << endl;
}