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;
long long i, n, p1, p2, v1, v2, k, a, b, d, pmin, pmax, vmin, vmax, pl, pu, vl,
    vu;
vector<long long> l;
int main() {
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  a = min(p1, v1);
  b = max(p2, v2);
  l.push_back(0);
  i = 0;
  while (l[l.size() - 1] <= b) {
    l.push_back(l[i] * 10 + 4);
    l.push_back(l[i] * 10 + 7);
    i++;
  }
  d = 0;
  for (i = 0; i < l.size() - k + 1; i++) {
    if (l[i] < a) continue;
    if (l[i + k - 1] > b) continue;
    pmin = max(p1, l[i - 1] + 1);
    pmax = min(p2, l[i]);
    vmin = max(v1, l[i + k - 1]);
    vmax = min(v2, l[i + k] - 1);
    if (pmax >= pmin && vmax >= vmin)
      d += (pmax - pmin + 1) * (vmax - vmin + 1);
    vmin = max(v1, l[i - 1] + 1);
    vmax = min(v2, l[i]);
    pmin = max(p1, l[i + k - 1]);
    pmax = min(p2, l[i + k] - 1);
    if (pmax >= pmin && vmax >= vmin)
      d += (pmax - pmin + 1) * (vmax - vmin + 1);
    if (k == 1 && l[i] <= vmax && l[i] >= vmin && l[i] >= pmin && l[i] <= pmax)
      d--;
  }
  double res = (double)d / (p2 - p1 + 1) / (v2 - v1 + 1);
  printf("%.9f", res);
}