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 pl, pr, vl, vr, l, r, rr, ll, ll1, rr1;
long long a[10000], sum;
long k, q, i, len, z, kol;
double otv;
void make(long ll, long long x, long m) {
  if (ll == m) {
    q++;
    a[q] = x;
    return;
  }
  make(ll + 1, x * 10 + 4, m);
  make(ll + 1, x * 10 + 7, m);
  return;
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  r = max(pr, vr);
  rr = r;
  len = 0;
  while (rr > 0) {
    len++;
    rr = rr / 10;
  }
  q = 0;
  for (i = 1; i <= len; i++) make(0, 0, i);
  a[q + 1] = 19999999999;
  a[0] = -1;
  sum = 0;
  kol = 0;
  l = pl;
  r = vr;
  for (i = 1; i <= q - k + 1; i++)
    if ((a[i] >= l) && (a[i + k - 1] <= r)) {
      z = 0;
      if ((a[i] >= pl) && (a[i] <= pr)) {
        rr = a[i];
        if (l > a[i - 1])
          ll = pl;
        else
          ll = a[i - 1] + 1;
      } else {
        if (a[i - 1] < pl) {
          ll = pl;
          rr = pr;
        } else if ((a[i - 1] >= pl) && (a[i - 1] < pr)) {
          ll = a[i - 1] + 1;
          rr = pr;
        } else
          z = 1;
      }
      if ((a[i + k - 1] >= vl) && (a[i + k - 1] <= vr)) {
        ll1 = a[i + k - 1];
        if (vr < a[i + k])
          rr1 = vr;
        else
          rr1 = a[i + k] - 1;
      } else {
        if (a[i + k] > vr) {
          ll1 = vl;
          rr1 = vr;
        } else if ((a[i + k] > vl) && (a[i + k] <= vr)) {
          ll1 = vl;
          rr1 = a[i + k] - 1;
        } else
          z = 1;
      }
      if (ll1 == rr) kol++;
      if (z == 0) sum += (rr - ll + 1) * (rr1 - ll1 + 1);
    }
  l = vl;
  r = pr;
  for (i = 1; i <= q - k + 1; i++)
    if ((a[i] >= l) && (a[i + k - 1] <= r)) {
      z = 0;
      if ((a[i] >= vl) && (a[i] <= vr)) {
        rr = a[i];
        if (l > a[i - 1])
          ll = vl;
        else
          ll = a[i - 1] + 1;
      } else {
        if (a[i - 1] < vl) {
          ll = vl;
          rr = vr;
        } else if ((a[i - 1] >= vl) && (a[i - 1] < vr)) {
          ll = a[i - 1] + 1;
          rr = vr;
        } else
          z = 1;
      }
      if ((a[i + k - 1] >= pl) && (a[i + k - 1] <= pr)) {
        ll1 = a[i + k - 1];
        if (pr < a[i + k])
          rr1 = pr;
        else
          rr1 = a[i + k] - 1;
      } else {
        if (a[i + k] > pr) {
          ll1 = pl;
          rr1 = pr;
        } else if ((a[i + k] > pl) && (a[i + k] <= pr)) {
          ll1 = pl;
          rr1 = a[i + k] - 1;
        } else
          z = 1;
      }
      if (z == 0) sum += (rr - ll + 1) * (rr1 - ll1 + 1);
    }
  sum = sum - kol;
  otv = ((double(sum) / double(pr - pl + 1)) *
         (double(sum) / double(vr - vl + 1)));
  if (sum > 0) otv = otv / double(sum);
  printf("%.9lf", otv);
  return 0;
}