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;
set<long long> st;
vector<long long> vec;
void calc(long long n, long long num) {
  if (n == 0) {
    st.insert(num);
    return;
  }
  calc(n - 1, num);
  calc(n - 1, num * 10 + 4);
  calc(n - 1, num * 10 + 7);
}
int main() {
  long long cs, t, i, j, k, x, y, z, ans, q, m, n, pl, pr, vl, vr, a, b;
  calc(11, 0);
  for (set<long long>::iterator sit = st.begin(); sit != st.end(); sit++)
    vec.push_back(*sit);
  while (cin >> pl >> pr >> vl >> vr >> k) {
    ans = 0;
    for (i = 1; i + k - 1 < vec.size(); i++) {
      x = vec[i];
      y = vec[i + k - 1];
      long long px = vec[i - 1] + 1;
      long long ny = vec[i + k] - 1;
      a = b = 0;
      a = min(x, pr) - max(px, pl) + 1;
      b = min(ny, vr) - max(y, vl) + 1;
      a = max(a, 0LL);
      b = max(b, 0LL);
      ans += a * b;
      a = b = 0;
      a = min(x, vr) - max(px, vl) + 1;
      b = min(ny, pr) - max(y, pl) + 1;
      a = max(a, 0LL);
      b = max(b, 0LL);
      ans += a * b;
    }
    if (k == 1) {
      for (i = 1; i < vec.size(); i++) {
        if (pl <= vec[i] && vec[i] <= pr && vl <= vec[i] && vec[i] <= vr) ans--;
      }
    }
    printf("%0.11f\n", (double)ans / (double)((pr - pl + 1) * (vr - vl + 1)));
  }
  return 0;
}