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 p1, p2, v1, v2, k, lk[100005], top, ans, all;
inline long long read() {
  long long ret = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -ff;
    ch = getchar();
  }
  while (isdigit(ch)) {
    ret = (ret << 3) + (ret << 1) + (ch ^ 48);
    ch = getchar();
  }
  return ret * ff;
}
void dfs(long long x, long long s) {
  lk[++top] = s;
  if (x == 10) return;
  dfs(x + 1, s * 10 + 4);
  dfs(x + 1, s * 10 + 7);
}
inline long long calc(long long x, long long y, long long a, long long b) {
  if (a >= x && b <= y) return b - a + 1;
  if (x >= a && y <= b) return y - x + 1;
  if (x <= a && a <= y && y <= b) return y - a + 1;
  if (a <= x && x <= b && b <= y) return b - x + 1;
  if (y <= a || b <= x) return 0;
}
signed main() {
  dfs(1, 0);
  sort(lk + 1, lk + top + 1);
  lk[0] = 0, lk[top + 1] = 2e9;
  p1 = read(), p2 = read(), v1 = read(), v2 = read(), k = read();
  all = (p2 - p1 + 1) * (v2 - v1 + 1);
  if (k == 1) {
    for (long long i = 1; i <= top; i++) {
      ans += calc(lk[i - 1] + 1, lk[i], p1, p2) *
             calc(lk[i], lk[i + 1] - 1, v1, v2);
      swap(p1, v1), swap(p2, v2);
      ans += calc(lk[i - 1] + 1, lk[i], p1, p2) *
             calc(lk[i], lk[i + 1] - 1, v1, v2);
      swap(p1, v1), swap(p2, v2);
      if (lk[i] >= p1 && lk[i] <= p2 && lk[i] >= v1 && lk[i] <= v2) ans--;
    }
    double res;
    res = (double)ans / (double)all;
    printf("%.12lf", res);
    return 0;
  }
  for (long long i = 1; i + k - 1 <= top; i++) {
    ans += calc(lk[i - 1] + 1, lk[i], p1, p2) *
           calc(lk[i + k - 1], lk[i + k] - 1, v1, v2);
    swap(p1, v1), swap(p2, v2);
    ans += calc(lk[i - 1] + 1, lk[i], p1, p2) *
           calc(lk[i + k - 1], lk[i + k] - 1, v1, v2);
    swap(p1, v1), swap(p2, v2);
  }
  double res;
  res = (double)ans / (double)all;
  printf("%.12lf", res);
  return 0;
}