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;
void generate(long long x, vector<int>& res) {
  if (x <= 1000000000) {
    if (x) res.push_back(x);
    long long a = 10LL * x + 4;
    long long b = 10LL * x + 7;
    generate(a, res);
    generate(b, res);
  }
}
vector<int> vec;
double prob(int a, int b, int l, int r) {
  int ca = max(a, l);
  int cb = min(b, r);
  if (cb < ca) return 0.0;
  return double(cb - ca + 1) / double(r - l + 1);
}
int main() {
  generate(0, vec);
  sort(vec.begin(), vec.end());
  int n = (int)vec.size();
  int pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  double res = 0;
  for (int i = 0; i < (n - k + 1); ++i) {
    int a = (i == 0) ? 1 : vec[i - 1] + 1;
    int b = vec[i];
    int c = vec[i + k - 1];
    int d = (i + k >= n) ? 1000000000 : vec[i + k] - 1;
    res += prob(a, b, vl, vr) * prob(c, d, pl, pr);
    res += prob(a, b, pl, pr) * prob(c, d, vl, vr);
    if (b == c && k == 1) res -= prob(b, b, vl, vr) * prob(c, c, pl, pr);
  }
  printf("%.9lf\n", res);
  return 0;
}