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;
#pragma comment(linker, "/STACK:64000000")
template <class T>
inline T sqr(T x) {
  return x * x;
}
vector<int> lucky;
void precalc() {
  lucky.clear();
  for (int len = 1; len <= 9; ++len) {
    for (int mask = 0; mask < (1 << len); ++mask) {
      int cur = 0;
      for (int i = 0; i < len; ++i) {
        if (mask & (1 << i))
          cur = cur * 10 + 4;
        else
          cur = cur * 10 + 7;
      }
      lucky.push_back(cur);
    }
  }
  sort((lucky).begin(), (lucky).end());
}
int l1, r1;
int l2, r2;
int n;
inline pair<int, int> intersect(pair<int, int> a, pair<int, int> b) {
  if (a.first > a.second || b.first > b.second) return pair<int, int>();
  int L = max(a.first, b.first);
  int R = min(a.second, b.second);
  return L <= R ? make_pair(L, R) : pair<int, int>();
}
int main() {
  precalc();
  cin >> l1 >> r1 >> l2 >> r2 >> n;
  double res = 0.0;
  for (int i = 0; i + n - 1 < (int)((lucky).size()); ++i) {
    int l = lucky[i];
    int r = lucky[i + n - 1];
    int left, right;
    pair<int, int> a, b;
    left = i ? max(lucky[i - 1] + 1, l1) : l1;
    right = i + n < (int)((lucky).size()) ? min(lucky[i + n] - 1, r2) : r2;
    a = make_pair(left, min(l, r1));
    b = make_pair(max(r, l2), right);
    left = max(min(l, r1) - left + 1, 0);
    right = max(right - max(r, l2) + 1, 0);
    res += left * 1ll * right;
    pair<int, int> c = intersect(a, b);
    left = i ? max(lucky[i - 1] + 1, l2) : l2;
    right = i + n < (int)((lucky).size()) ? min(lucky[i + n] - 1, r1) : r1;
    a = make_pair(left, min(l, r2));
    b = make_pair(max(r, l1), right);
    left = max(min(l, r2) - left + 1, 0);
    right = max(right - max(r, l1) + 1, 0);
    res += left * 1ll * right;
    c = intersect(c, intersect(a, b));
    if (c.first) {
      res -= c.second - c.first + 1;
    }
  }
  res /= (r1 - l1 + 1);
  res /= (r2 - l2 + 1);
  printf("%.15lf\n", res);
  return 0;
}