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;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = ((ans % m) * (a % m)) % m;
    b /= 2;
    a = ((a % m) * (a % m)) % m;
  }
  return ans % m;
}
const long double eps = 1e-10L;
vector<long long> v;
long long pl, pr, vl, vr, k;
void generate(long long id) {
  if (id > 10000000009) return;
  v.push_back(id);
  generate(id * 10 + 4);
  generate(id * 10 + 7);
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  v.push_back(0);
  generate(4);
  generate(7);
  sort(v.begin(), v.end());
  double ans = 0.00;
  if (k == 1) {
    for (long long i = 1; i <= (long long)v.size() - k; i++) {
      long long x = v[i];
      long long y = v[i + k - 1];
      if (pl <= x && vr >= y && pr > v[i - 1] && vl < v[i + k]) {
        ans += abs((max(v[i - 1] + 1, pl) - min(x, pr) - 1) *
                   (max(y, vl) - min(v[i + k] - 1, vr) - 1));
      }
      if (vl <= x && pr >= y && vr > v[i - 1] && pl < v[i + k]) {
        ans += abs((max(v[i - 1] + 1, vl) - min(x, vr) - 1) *
                   (max(y, pl) - min(v[i + k] - 1, pr) - 1));
      }
    }
    for (int i = 1; i < v.size(); i++) {
      if (v[i] <= pr && v[i] >= pl)
        if (v[i] <= vr && v[i] >= vl) ans--;
    }
    double deno = (double)(pr - pl + 1) * (vr - vl + 1);
    ans /= (double)deno;
    cout << fixed << setprecision(12) << ans;
    return 0;
  }
  for (long long i = 1; i <= (long long)v.size() - k; i++) {
    long long x = v[i];
    long long y = v[i + k - 1];
    if (pl <= x && vr >= y && pr > v[i - 1] && vl < v[i + k]) {
      ans += abs((max(v[i - 1] + 1, pl) - min(x, pr) - 1) *
                 (max(y, vl) - min(v[i + k] - 1, vr) - 1));
    }
    if (vl <= x && pr >= y && vr > v[i - 1] && pl < v[i + k]) {
      ans += abs((max(v[i - 1] + 1, vl) - min(x, vr) - 1) *
                 (max(y, pl) - min(v[i + k] - 1, pr) - 1));
    }
  }
  double deno = (double)(pr - pl + 1) * (vr - vl + 1);
  ans /= (double)deno;
  cout << fixed << setprecision(12) << ans;
}