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, m, p1, p2, v1, v2, k, num;
int rec[2005];
void dfs(long long val) {
  if (val > 1000000000) return;
  rec[num++] = val;
  dfs(val * 10 + 4);
  dfs(val * 10 + 7);
}
void solve() {
  int i, len1, len2, s, t, flag;
  double ans, sum;
  sort(rec + 1, rec + num);
  rec[num] = 0x3fffffff;
  sum = (long long)(v2 - v1 + 1) * (p2 - p1 + 1);
  ans = 0;
  for (i = 1; i <= num - k; i++) {
    flag = 0;
    s = rec[i];
    t = rec[i + k - 1];
    if (s >= v1 && t <= p2) {
      len1 = ((((s - v1) > (v2 - v1) ? (v2 - v1) : (s - v1))) >
                      (((v2 - rec[i - 1] - 1) > (s - rec[i - 1] - 1)
                            ? (s - rec[i - 1] - 1)
                            : (v2 - rec[i - 1] - 1)))
                  ? (((v2 - rec[i - 1] - 1) > (s - rec[i - 1] - 1)
                          ? (s - rec[i - 1] - 1)
                          : (v2 - rec[i - 1] - 1)))
                  : (((s - v1) > (v2 - v1) ? (v2 - v1) : (s - v1)))) +
             1;
      len2 = ((((p2 - t) > (p2 - p1) ? (p2 - p1) : (p2 - t))) >
                      (((rec[i + k] - p1 - 1) > (rec[i + k] - t - 1)
                            ? (rec[i + k] - t - 1)
                            : (rec[i + k] - p1 - 1)))
                  ? (((rec[i + k] - p1 - 1) > (rec[i + k] - t - 1)
                          ? (rec[i + k] - t - 1)
                          : (rec[i + k] - p1 - 1)))
                  : (((p2 - t) > (p2 - p1) ? (p2 - p1) : (p2 - t)))) +
             1;
      if (len1 > 0 && len2 > 0) ans += (long long)len1 * len2;
      flag = 1;
    }
    if (s >= p1 && t <= v2) {
      len1 = ((((s - p1) > (p2 - p1) ? (p2 - p1) : (s - p1))) >
                      (((p2 - rec[i - 1] - 1) > (s - rec[i - 1] - 1)
                            ? (s - rec[i - 1] - 1)
                            : (p2 - rec[i - 1] - 1)))
                  ? (((p2 - rec[i - 1] - 1) > (s - rec[i - 1] - 1)
                          ? (s - rec[i - 1] - 1)
                          : (p2 - rec[i - 1] - 1)))
                  : (((s - p1) > (p2 - p1) ? (p2 - p1) : (s - p1)))) +
             1;
      len2 = ((((v2 - t) > (v2 - v1) ? (v2 - v1) : (v2 - t))) >
                      (((rec[i + k] - v1 - 1) > (rec[i + k] - t - 1)
                            ? (rec[i + k] - t - 1)
                            : (rec[i + k] - v1 - 1)))
                  ? (((rec[i + k] - v1 - 1) > (rec[i + k] - t - 1)
                          ? (rec[i + k] - t - 1)
                          : (rec[i + k] - v1 - 1)))
                  : (((v2 - t) > (v2 - v1) ? (v2 - v1) : (v2 - t)))) +
             1;
      if (len1 > 0 && len2 > 0) ans += (long long)len1 * len2;
      if (k == 1 && flag && p1 <= rec[i] && p2 >= rec[i] && v1 <= rec[i] &&
          v2 >= rec[i])
        ans--;
    }
  }
  ans /= sum;
  printf("%.10lf\n", ans);
}
int main() {
  dfs(0);
  scanf("%d%d%d%d%d", &p1, &p2, &v1, &v2, &k);
  solve();
  return 0;
}