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 di[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dj[] = {1, -1, 0, 0, -1, 1, 1, -1};
vector<int> lucky;
int intersect(int l, int r, int s, int e) {
  return max(0, min(r, e) - max(l, s) + 1);
}
int main() {
  int s1, e1, s2, e2, k;
  scanf("%d%d%d%d%d", &s1, &e1, &s2, &e2, &k);
  for (int i = 1; i <= 9; i++) {
    for (int j = 0; j < (1 << i); j++) {
      int num = 0;
      for (int k = 0; k < i; k++) {
        num *= 10;
        num += (j & (1 << k)) ? 7 : 4;
      }
      lucky.push_back(num);
    }
  }
  lucky.push_back(0);
  lucky.push_back(1e9 + 1);
  sort((lucky).begin(), (lucky).end());
  double ans = 0;
  int r = k - 1;
  while (++r < (int)lucky.size() - 1) {
    int l = r - k + 1;
    int a = intersect(s1, e1, lucky[l - 1] + 1, lucky[l]);
    int b = intersect(s2, e2, lucky[r], lucky[r + 1] - 1);
    ans += (a / (e1 - s1 + 1.0)) * (b / (e2 - s2 + 1.0));
    a = intersect(s2, e2, lucky[l - 1] + 1, lucky[l]);
    b = intersect(s1, e1, lucky[r], lucky[r + 1] - 1);
    ans += (a / (e2 - s2 + 1.0)) * (b / (e1 - s1 + 1.0));
    if (l == r && lucky[r] >= s1 && lucky[r] <= e1 && lucky[r] >= s2 &&
        lucky[r] <= e2)
      ans -= (1 / (e1 - s1 + 1.0)) * (1 / (e2 - s2 + 1.0));
  }
  printf("%.10f", ans);
}