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 arr[1044], cnt, a1, a2, b1, b2, k, x, y;
void generate_lucky_number(long long nm) {
  if (nm > y) return;
  if (nm >= x) {
    arr[cnt++] = nm;
  }
  generate_lucky_number(nm * 10 + 4);
  generate_lucky_number(nm * 10 + 7);
}
double calcl(long long p, long long l, long long r) {
  long long tmp;
  if (arr[p] < l) return 0;
  if (arr[p] > r) {
    if (p == 0 || arr[p - 1] < l) return (double)(r - l + 1);
    if (arr[p - 1] > r) return 0;
    return (double)(r - arr[p - 1]);
  }
  if (p != 0) {
    if (arr[p - 1] >= l) return (double)(arr[p] - arr[p - 1]);
  }
  return (double)(arr[p] - l + 1);
}
double calch(long long p, long long l, long long r) {
  if (arr[p] > r) return 0;
  if (arr[p] < l) {
    if (p == cnt - 1 || arr[p + 1] > r) return (double)(r - l + 1);
    if (arr[p + 1] < l) return 0;
    return (double)(arr[p + 1] - l);
  }
  if (p != cnt - 1) {
    if (arr[p + 1] <= r) return (double)(arr[p + 1] - arr[p]);
  }
  return (double)(r - arr[p] + 1);
}
int main() {
  long long i, j;
  double total, sol, tmp1, tmp2;
  scanf("%lld%lld%lld%lld%lld", &a1, &a2, &b1, &b2, &k);
  total = (double)(a2 - a1 + 1);
  total *= (double)(b2 - b1 + 1);
  x = min(a1, b1);
  y = max(a2, b2);
  generate_lucky_number(0);
  sort(arr, arr + cnt);
  sol = 0;
  for (i = 0; i < cnt; i++) {
    j = i + k - 1;
    if (j >= cnt) break;
    tmp1 = calcl(i, a1, a2);
    tmp1 *= calch(j, b1, b2);
    sol += tmp1;
    tmp2 = calcl(i, b1, b2);
    tmp2 *= calch(j, a1, a2);
    sol += tmp2;
    if (k == 1 && tmp1 > 0 && tmp2 > 0) sol = sol - 1;
  }
  printf("%.12lf\n", sol / total);
  return 0;
}