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<long long> v;
void Generate(long long x) {
  if (x > 1e9) return;
  v.push_back(x);
  Generate(x * 10 + 4);
  Generate(x * 10 + 7);
}
long long Intersect(int a, int b, int l, int r) {
  if (r < a || b < l) return 0;
  return min(b, r) - max(l, a) + 1;
}
int main() {
  v.push_back(1e10);
  Generate(0);
  sort(v.begin(), v.end());
  double ans = 0;
  long long L[2], R[2], k;
  cin >> L[0] >> R[0] >> L[1] >> R[1] >> k;
  for (int i = k; i + 1 < v.size(); i++) {
    for (int flag = 0; flag < 2; flag++)
      ans += 1.0 * Intersect(v[i], v[i + 1] - 1, L[flag], R[flag]) /
             (R[flag] - L[flag] + 1) * 1.0 *
             Intersect(v[i - k] + 1, v[i - k + 1], L[!flag], R[!flag]) /
             (R[!flag] - L[!flag] + 1);
    if (k == 1 && L[0] <= v[i] && v[i] <= R[0] && L[1] <= v[i] && v[i] <= R[1])
      ans -= 1.0 / ((R[0] - L[0] + 1) * (R[1] - L[1] + 1));
  }
  printf("%.10f\n", ans);
  return 0;
}