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 q[11], c[10][10], p2[10];
void change(long long& a, long long& b) {
  a = a ^ b;
  b = a ^ b;
  a = a ^ b;
}
long long min(long long a, long long b) { return a < b ? a : b; }
long long max(long long a, long long b) { return a > b ? a : b; }
long long f(long long n0) {
  if (n0 == 0) return 0;
  long long i, j, k, t, n = n0, ans = 0;
  q[0] = 0;
  while (n > 0) {
    q[++q[0]] = n % 10;
    n /= 10;
  }
  for (i = 1; i <= q[0] / 2; i++) change(q[i], q[q[0] + 1 - i]);
  ans = p2[q[0]] - 2;
  t = 1;
  for (i = 1; i <= q[0]; i++) {
    if (q[i] > 7) {
      ans += p2[q[0] - i + 1];
      break;
    } else if (q[i] > 4) {
      ans += p2[q[0] - i];
      if (q[i] < 7) break;
    } else if (q[i] < 4)
      break;
  }
  if (i > q[0]) ans += t;
  return ans;
}
long long bs(long long t) {
  long long l, r, mid;
  l = 1;
  r = 1000000010;
  while (l < r - 1) {
    mid = (l + r) / 2;
    if (f(mid) < t)
      l = mid;
    else
      r = mid;
  }
  if (f(r) == t && f(l) == t - 1)
    return r;
  else
    return l;
}
int main() {
  long long i, j, pl, pr, vl, vr, k, a, b, d, e;
  long long ans;
  double as;
  for (i = 0; i < 10; i++) {
    p2[i] = (1 << i);
    c[i][0] = 1;
    c[i][i] = 1;
    for (j = 1; j < i; j++) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
  while (cin >> pl >> pr >> vl >> vr >> k) {
    ans = 0;
    a = pl;
    b = bs(f(a) + 1);
    if (f(a) == f(a - 1))
      a--;
    else {
      d = min(bs(f(a) + k), vr + 1) - max(bs(f(a) + k - 1), vl);
      if (d > 0) ans += d;
    }
    while (b <= pr) {
      d = min(bs(f(a) + k + 1), vr + 1) - max(bs(f(a) + k), vl);
      if (d > 0) ans += (b - a) * d;
      a = b;
      b = bs(f(b) + 1);
    }
    d = min(bs(f(a) + k + 1), vr + 1) - max(bs(f(a) + k), vl);
    if (d > 0) ans += (pr - a) * d;
    a = vl;
    b = bs(f(a) + 1);
    if (f(a) == f(a - 1))
      a--;
    else {
      d = min(bs(f(a) + k), pr + 1) - max(bs(f(a) + k - 1), pl);
      if (d > 0) ans += d;
    }
    while (b <= vr) {
      d = min(bs(f(a) + k + 1), pr + 1) - max(bs(f(a) + k), pl);
      if (d > 0) ans += (b - a) * d;
      a = b;
      b = bs(f(b) + 1);
    }
    d = min(bs(f(a) + k + 1), pr + 1) - max(bs(f(a) + k), pl);
    if (d > 0) ans += (vr - a) * d;
    if (k == 1 && min(pr, vr) - max(pl, vl) >= 0) {
      ans = ans - (f(min(pr, vr)) - f(max(pl, vl)));
      if (f(max(pl, vl)) != f(max(pl, vl) - 1)) ans--;
    }
    as = (double)ans / (vr - vl + 1) / (pr - pl + 1);
    printf("%.10lf\n", as);
  }
  return 0;
}