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;
int a[11111], n = 0, l1, r1, l2, r2, pl, pr, vl, vr, k;
long long prob = 0;
void rec(int x, int k) {
  a[++n] = x;
  if (k == 9) return;
  rec(x * 10 + 4, k + 1);
  rec(x * 10 + 7, k + 1);
}
long long f(int a, int b, int c, int d) {
  a = ((a) > (c) ? (a) : (c));
  b = ((b) < (d) ? (b) : (d));
  if (a > b) return 0;
  return b - a + 1;
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  rec(4, 1);
  rec(7, 1);
  a[0] = 0;
  a[n + 1] = INT_MAX;
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n - k + 1; i++) {
    l1 = a[i - 1] + 1, r1 = a[i];
    l2 = a[i + k - 1], r2 = a[i + k] - 1;
    prob += f(l1, r1, pl, pr) * f(l2, r2, vl, vr);
    prob += f(l1, r1, vl, vr) * f(l2, r2, pl, pr);
    prob -= (k == 1 && a[i] >= pl && a[i] <= pr && a[i] >= vl && a[i] <= vr);
  }
  cout.precision(13);
  cout << fixed << (double)prob / (pr - pl + 1) / (vr - vl + 1) << endl;
  return 0;
}