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;
priority_queue<int> QwQ;
int l = 2;
int c[1005], a[1005][1005];
long long max(long long b, long long c) {
  if (b > c) return b;
  return c;
}
long long min(long long b, long long c) {
  if (b > c) return c;
  return b;
}
int main() {
  int i, j, k, f = 0, t, p, q;
  long long x1, x2, y1, y2, t1, t2, ttt;
  double ans;
  long long Ans1, Ans2, Ans = 0, m1, m2, sum;
  int n;
  k = 1;
  t = 2;
  c[1] = 4, c[2] = 7;
  for (i = 2; i <= 9; i++) {
    c[++l] = c[k] * 10 + 4;
    p = l;
    c[++l] = c[k] * 10 + 7;
    for (j = k + 1; j <= t; j++) {
      c[++l] = c[j] * 10 + 4;
      c[++l] = c[j] * 10 + 7;
    }
    t = l;
    k = p;
  }
  scanf("%I64d%I64d%I64d%I64d", &x1, &y1, &x2, &y2);
  t1 = y1 - x1 + 1;
  t2 = y2 - x2 + 1;
  scanf("%d", &n);
  c[0] = 0;
  c[1023] = 2000000009;
  for (i = 1; i <= 1022; i++) {
    m1 = max(c[i - 1] + 1, x1);
    m2 = min(c[i], y1);
    Ans1 = m2 - m1 + 1;
    if (Ans1 <= 0) continue;
    j = i + n - 1;
    if (j > 1022) break;
    m1 = max(c[j], x2);
    m2 = min(c[j + 1] - 1, y2);
    Ans2 = m2 - m1 + 1;
    if (Ans2 <= 0) continue;
    Ans += Ans1 * Ans2;
  }
  for (i = 1; i <= 1022; i++) {
    m1 = max(c[i - 1] + 1, x2);
    m2 = min(c[i], y2);
    Ans1 = m2 - m1 + 1;
    if (Ans1 <= 0) continue;
    j = i + n - 1;
    if (j > 1022) break;
    m1 = max(x1, c[j]);
    m2 = min(c[j + 1] - 1, y1);
    Ans2 = m2 - m1 + 1;
    if (Ans2 <= 0) continue;
    Ans += Ans1 * Ans2;
  }
  if (n == 1) {
    for (i = 1; i <= 1022; i++) {
      if (x1 <= c[i] && c[i] <= y1 && x2 <= c[i] && c[i] <= y2) Ans--;
    }
  }
  sum = t1 * t2;
  ans = (double)(Ans * 1.0) / (double)(sum * 1.0);
  printf("%.12f\n", ans);
  return 0;
}