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<int> s;
const int dig[] = {0, 4, 7};
int num;
void init(int x, int p) {
  if (x == 9) {
    if (num != 0) s.push_back(num);
    return;
  }
  int b = 0;
  if (num != 0) b = 1;
  for (int i = b; i < 3; ++i) {
    num += p * dig[i];
    init(x + 1, p / 10);
    num -= p * dig[i];
  }
}
int k;
long long solve(int vl, int vr, int pl, int pr) {
  long long sum = 0;
  for (int i = vl; i <= vr;) {
    int p = lower_bound(s.begin(), s.end(), i) - s.begin();
    if (p == s.size()) break;
    long long t = 0;
    if (p + k - 1 < s.size()) {
      if (pl <= s[p + k - 1] && s[p + k - 1] <= pr) {
        if (p + k == s.size())
          t = pr - s[p + k - 1] + 1;
        else
          t = min(s[p + k] - 1, pr) - s[p + k - 1] + 1;
      } else if (pl > s[p + k - 1]) {
        if (p + k < s.size() && pl < s[p + k])
          t = min(s[p + k] - 1, pr) - pl + 1;
        else if (p + k == s.size())
          t = pr - pl + 1;
      }
    }
    sum += t * (min(vr, s[p]) - i + 1);
    i = s[p] + 1;
  }
  return sum;
}
int main() {
  init(0, 100000000);
  int vl, vr, pl, pr;
  long long n1, n2;
  scanf("%d%d%d%d%d", &vl, &vr, &pl, &pr, &k);
  n2 = (vr - vl + 1) * (long long)(pr - pl + 1);
  n1 = solve(vl, vr, pl, pr) + solve(pl, pr, vl, vr);
  int l = max(vl, pl), r = min(vr, pr);
  if (k == 1 && l <= r)
    n1 -=
        upper_bound(s.begin(), s.end(), r) - lower_bound(s.begin(), s.end(), l);
  cout << setprecision(12) << fixed << (long double)n1 / (long double)n2
       << endl;
  return 0;
}