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;
inline long long read() {
  long long x = 0, w = 0;
  char ch = getchar();
  while (!isdigit(ch)) w |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
  return w ? -x : x;
}
long long ans, a, b, c, d, k, ok[999999], tot;
void dfs(long long k, long long t) {
  ok[++tot] = t;
  if (t * 10 + 4 > 1e9) return;
  dfs(k + 1, t * 10 + 4);
  dfs(k + 1, t * 10 + 7);
}
long long calc(long long l, long long r, long long x, long long y) {
  long long L = max(l, x), R = min(y, r);
  return max(R - L + 1, 0ll);
}
signed main() {
  a = read(), b = read(), c = read(), d = read(), k = read();
  tot = -1;
  dfs(0, 0);
  ok[++tot] = 0x3f3f3f3f;
  sort(ok + 1, ok + 1 + tot);
  for (long long i = 1; i + k <= tot; ++i)
    ans += calc(a, b, ok[i - 1] + 1, ok[i]) *
           calc(c, d, ok[i + k - 1], ok[i + k] - 1);
  for (long long i = 1; i + k <= tot; ++i)
    ans += calc(c, d, ok[i - 1] + 1, ok[i]) *
           calc(a, b, ok[i + k - 1], ok[i + k] - 1);
  if (k == 1)
    for (long long i = 1; i < tot; ++i)
      if (calc(a, b, ok[i], ok[i]) && calc(c, d, ok[i], ok[i])) --ans;
  printf("%.15lf", 1.0 * ans / ((d - c + 1) * (b - a + 1)));
  return 0;
}