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;
int n, k;
int a[1200];
void gen(int x, int y) {
  a[n++] = x;
  if (y == 9) return;
  gen(x * 10 + 4, y + 1);
  gen(x * 10 + 7, y + 1);
}
long long cross(int l1, int r1, int l2, int r2) {
  int l, r;
  l = max(l1, l2);
  r = min(r1, r2);
  if (l > r)
    return 0;
  else
    return (long long)(r - l + 1);
}
long long cal(int l1, int r1, int l2, int r2) {
  int i;
  long long v1, v2, res = 0;
  for (i = 1; i + k < n; i++) {
    v1 = cross(a[i - 1] + 1, a[i], l1, r1);
    v2 = cross(a[i + k - 1], a[i + k] - 1, l2, r2);
    long long temp = v1 * v2;
    res += temp;
  }
  return res;
}
int main() {
  gen(0, 0);
  sort(a, a + n);
  a[n++] = 1000000001;
  int l1, r1, l2, r2;
  scanf("%d%d%d%d%d", &l1, &r1, &l2, &r2, &k);
  long long y = (long long)(r1 - l1 + 1) * (long long)(r2 - l2 + 1);
  long long x = cal(l1, r1, l2, r2) + cal(l2, r2, l1, r1);
  if (k == 1)
    for (int i = 1; i < n - 1; i++)
      if (l1 <= a[i] && a[i] <= r1 && l2 <= a[i] && a[i] <= r2) x--;
  printf("%.15f\n", (double)x / (double)y);
  return 0;
}