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 a[2000], cnt;
long long pl, pr, vl, vr, K;
void dfs(long long x) {
  if (x > 1000000000LL) return;
  a[++cnt] = x;
  dfs(x * 10LL + 4);
  dfs(x * 10LL + 7);
}
long long ok(long long a1, long long a2, long long b1, long long b2) {
  long long l = max(a1, b1), r = min(a2, b2);
  if (l <= r)
    return r - l + 1LL;
  else
    return 0;
}
int main() {
  int i, j, k;
  a[0] = 0;
  cnt = 0;
  dfs(4);
  dfs(7);
  sort(a, a + cnt + 1);
  a[cnt + 1] = 1000000009LL;
  while (scanf("%lld%lld%lld%lld%lld", &pl, &pr, &vl, &vr, &K) != EOF) {
    long long tot = 0;
    for (i = 1; i + K - 1 <= cnt; i++) {
      long long x, y;
      x = ok(a[i - 1] + 1, a[i], pl, pr) *
          ok(a[i + K - 1], a[i + K] - 1, vl, vr);
      y = ok(a[i - 1] + 1, a[i], vl, vr) *
          ok(a[i + K - 1], a[i + K] - 1, pl, pr);
      tot += x + y;
      if (x && y && K == 1) tot--;
    }
    printf("%.12f\n",
           (double)tot / ((pr - pl + 1) * 1.0) / ((vr - vl + 1) * 1.0));
  }
  return 0;
}