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>
int an, arr[100000];
int l1, r1, l2, r2, k;
const int inf = 1000000001;
void dfs(int d, int val) {
  arr[an++] = val;
  if (d == 9) return;
  dfs(d + 1, val * 10 + 4);
  dfs(d + 1, val * 10 + 7);
}
inline void gen() {
  an = 0;
  dfs(0, 0);
  std::sort(arr, arr + an);
}
inline int min(int a, int b) { return a < b ? a : b; }
inline int max(int a, int b) { return a > b ? a : b; }
inline bool bet(int x, int l, int r) { return x >= l && x <= r; }
inline int intersect(int l1, int r1, int l2, int r2) {
  return max(min(r1, r2) - max(l1, l2) + 1, 0);
}
inline double solve() {
  int i, j, a, b, ll, rr, lb, rb;
  long long cc = 0;
  long long dd = (long long)(r1 - l1 + 1) * (r2 - l2 + 1);
  for (i = 0; i < an; i++) {
    j = i + k - 1;
    if (j >= an) break;
    if (arr[j] > max(r1, r2)) break;
    cc += (long long)intersect(i ? arr[i - 1] + 1 : 1, arr[i], l1, r1) *
          intersect(arr[j], j < an - 1 ? arr[j + 1] - 1 : inf, l2, r2);
    cc += (long long)intersect(i ? arr[i - 1] + 1 : 1, arr[i], l2, r2) *
          intersect(arr[j], j < an - 1 ? arr[j + 1] - 1 : inf, l1, r1);
  }
  if (k == 1) {
    for (i = 0; i < an; i++)
      if (bet(arr[i], l1, r1) && bet(arr[i], l2, r2)) cc--;
  }
  return (double)cc / dd;
}
int main(void) {
  int i;
  gen();
  while (scanf("%d %d %d %d %d", &l1, &r1, &l2, &r2, &k) == 5) {
    printf("%.12lf\n", solve());
  }
  return 0;
}