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;
const int MOD(1000000007);
const int INF((1 << 30) - 1);
const int MAXN();
long long intersect(long long x1, long long y1, long long x2, long long y2) {
  return max(0ll, min(y1, y2) - max(x1, x2) + 1);
}
int main() {
  vector<long long> a;
  a.push_back(4);
  a.push_back(7);
  for (int i = 0; i < a.size(); i++) {
    if (a[i] * 10 + 4 <= 1e9) a.push_back(a[i] * 10 + 4);
    if (a[i] * 10 + 7 <= 1e9) a.push_back(a[i] * 10 + 7);
  }
  int k;
  long long x1, y1, x2, y2;
  cin >> x1 >> y1 >> x2 >> y2 >> k;
  long long ans = 0, l, r;
  for (int i = k - 1; i <= a.size() - 1; i++) {
    if (i - k >= 0)
      l = a[i - k] + 1;
    else
      l = 1;
    if (i + 1 < a.size())
      r = a[i + 1] - 1;
    else
      r = 1e9;
    ans += intersect(l, a[i - k + 1], x1, y1) * intersect(a[i], r, x2, y2);
    ans += intersect(l, a[i - k + 1], x2, y2) * intersect(a[i], r, x1, y1);
    if (k == 1)
      ans -= intersect(a[i], a[i], x1, y1) * intersect(a[i], a[i], x2, y2);
  }
  printf("%.10lf", ans * 1.0 / ((y1 - x1 + 1) * (y2 - x2 + 1)));
}