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;
const int N = 3e5 + 9, inf = 1e18, mod = 1e9 + 7;
vector<int> prime_Numbers;
bool primes[1000005];
double dot(complex<double> a, complex<double> b) {
  return (conj(a) * b).real();
}
double cross(complex<double> a, complex<double> b) {
  return (conj(a) * b).imag();
}
vector<long long> v;
void solve(long long n) {
  if (n > 1e10) return;
  v.push_back(n);
  solve(n * 10 + 4);
  solve(n * 10 + 7);
}
long long calc(long long left, long long right, long long grblft,
               long long grbright) {
  return max(0ll, min(right, grbright) - max(left, grblft) + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    long long pl, pr, vl, vr, k;
    cin >> pl >> pr >> vl >> vr >> k;
    solve(0);
    sort(v.begin(), v.end());
    long long ans = 0;
    for (int i = 1; i + k < v.size(); i++) {
      long long op1 = calc(pl, pr, v[i - 1] + 1, v[i]);
      long long op2 = calc(vl, vr, v[i + k - 1], v[i + k] - 1);
      ans += op1 * op2;
      op1 = calc(vl, vr, v[i - 1] + 1, v[i]);
      op2 = calc(pl, pr, v[i + k - 1], v[i + k] - 1);
      ans += op1 * op2;
      if (k == 1) {
        if (v[i] <= pr && v[i] >= pl && v[i] <= vr && v[i] >= vl) ans--;
      }
    }
    cout << fixed << setprecision(12)
         << (long double)ans / (long double)((pr - pl + 1) * (vr - vl + 1));
  }
  return 0;
}