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 l1, l2, r1, r2;
vector<long long> v;
void f(long long x) {
  if (x > r1 && x > r2) return;
  if (x >= min(l1, l2) && x <= max(r1, r2)) v.push_back(x);
  f(x * 10 + 7);
  f(x * 10 + 4);
}
int main() {
  long long n, k, i, mn, mx;
  double db = 0;
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  f(0);
  sort(v.begin(), v.end());
  long long l = 0, r = k - 1;
  double ans = 0;
  for (r = k - 1; r < v.size(); r++) {
    long long x = min(v[l], r1);
    long long y = max(v[r], l2);
    long long mn, mx, q1 = 0, q2 = 0;
    if (l == 0)
      mn = l1;
    else
      mn = max(l1, v[l - 1] + 1);
    if (r == v.size() - 1)
      mx = r2;
    else
      mx = min(v[r + 1] - 1, r2);
    if ((mx - y + 1) > 0 && (x - mn + 1) > 0)
      ans += (1.0 * ((x - mn + 1) * (mx - y + 1))) /
             (1.0 * (r1 - l1 + 1) * (r2 - l2 + 1)),
          q1 = 1;
    x = max(v[r], l1);
    y = min(v[l], r2);
    if (l == 0)
      mn = l2;
    else
      mn = max(l2, v[l - 1] + 1);
    if (r == v.size() - 1)
      mx = r1;
    else
      mx = min(v[r + 1] - 1, r1);
    if ((y - mn + 1) > 0 && (mx - x + 1) > 0)
      ans += (1.0 * (y - mn + 1) * (mx - x + 1)) /
             (1.0 * (r1 - l1 + 1) * (r2 - l2 + 1)),
          q2 = 1;
    if (q1 == 1 && q2 == 1 && k == 1)
      ans -= (1.0 / (1.0 * (r1 - l1 + 1) * (r2 - l2 + 1)));
    l++;
  }
  printf("%.15f", min(1.0, ans));
}