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 INF = 1e9 + 10;
bool ch(int a) {
  while (a > 0) {
    if (a % 10 != 4 && a % 10 != 7) return false;
    a /= 10;
  }
  return true;
}
int a[2000], kol = 0;
long long schetl(int l, int r, int d) {
  if (l > a[d]) return 0;
  if (r < a[d - 1]) return 0;
  int k;
  k = min(a[d], r) - max(a[d - 1] + 1, l);
  if (k >= 0)
    return k + 1;
  else
    return 0;
}
long long schetr(int l, int r, int d) {
  if (r < a[d]) return 0;
  if (l > a[d + 1]) return 0;
  int k;
  k = min(a[d + 1] - 1, r) - max(a[d], l);
  if (k >= 0)
    return k + 1;
  else
    return 0;
}
int main() {
  int i, uk = 1, k;
  long long res = 0, x, y, sum, d;
  long long pl, pr, vl, vr;
  long double z, res1;
  cin >> pl >> pr >> vl >> vr >> k;
  for (i = 0; i < 1024; i++) {
    x = i;
    d = 0;
    while (x > 1) {
      d *= 10;
      if (x % 2 == 0)
        d += 4;
      else
        d += 7;
      x /= 2;
    }
    a[kol] = d;
    kol++;
  }
  a[kol] = INF;
  kol++;
  sort(a, a + kol);
  while (uk + k < kol) {
    x = schetl(pl, pr, uk);
    y = schetr(vl, vr, uk + k - 1);
    res += x * y;
    x = schetl(vl, vr, uk);
    y = schetr(pl, pr, uk + k - 1);
    res += x * y;
    uk++;
  }
  for (i = 0; i < kol - 1; i++)
    if (a[i] >= pl && a[i] <= pr && a[i] >= vl && a[i] <= vr && k == 1) res--;
  sum = (1 + pr - pl) * (vr - vl + 1);
  cout.precision(15);
  res1 = res;
  z = res1 / sum;
  cout << fixed << z;
  return 0;
}