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 pl, pr, vl, vr, k;
vector<int> lucky;
void gen_lucky(int len) {
  for (int i = 0; i < (1 << len); i++) {
    int num = 0;
    for (int j = 0; j < len; j++) {
      if (i & (1 << j))
        num = num * 10 + 4;
      else
        num = num * 10 + 7;
    }
    lucky.push_back(num);
  }
}
double prob(int l1, int r1, int l2, int r2) {
  return max(0, min(r1, r2) - max(l1, l2) + 1) / (r2 - l2 + 1.0);
}
int main() {
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k);
  for (int i = 1; i < 10; i++) gen_lucky(i);
  lucky.push_back(0);
  lucky.push_back(0x3f3f3f3f);
  sort(lucky.begin(), lucky.end());
  double ans = 0.0;
  for (int i = 0; i + k + 1 < lucky.size(); i++) {
    ans += prob(lucky[i] + 1, lucky[i + 1], pl, pr) *
           prob(lucky[i + k], lucky[i + k + 1] - 1, vl, vr);
    ans += prob(lucky[i] + 1, lucky[i + 1], vl, vr) *
           prob(lucky[i + k], lucky[i + k + 1] - 1, pl, pr);
    if (k == 1)
      ans -= prob(lucky[i + 1], lucky[i + 1], pl, pr) *
             prob(lucky[i + 1], lucky[i + 1], vl, vr);
  }
  printf("%.12lf\n", ans);
}