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;
void mytimer(string task) {}
void ext(int c) {}
int a1, a2, b1, b2;
int k;
vector<int> lucky;
void gen(int a, int l) {
  if (l == 10) return;
  if (a) lucky.push_back(a);
  gen(a * 10 + 4, l + 1);
  gen(a * 10 + 7, l + 1);
}
double doit1(int v1, int v2) {
  v1 = max(v1, a1);
  v2 = min(v2, a2);
  if (v2 < v1) return 0;
  return 1. * (v2 - v1 + 1) / (a2 - a1 + 1);
}
double doit2(int v1, int v2) {
  v1 = max(v1, b1);
  v2 = min(v2, b2);
  if (v2 < v1) return 0;
  return 1. * (v2 - v1 + 1) / (b2 - b1 + 1);
}
int main() {
  cin >> a1 >> a2 >> b1 >> b2 >> k;
  gen(0, 0);
  lucky.push_back(1000000000 + 100);
  lucky.push_back(-100);
  sort(lucky.begin(), lucky.end());
  double res = 0;
  for (int i = 1; i + k < lucky.size(); ++i) {
    int p1 = lucky[i - 1] + 1;
    int p2 = lucky[i];
    int q1 = lucky[i + k - 1];
    int q2 = lucky[i + k] - 1;
    if (p2 == q1) {
      res += doit1(p1, p2) * doit2(q1 + 1, q2) +
             doit2(p1, p2) * doit1(q1 + 1, q2) +
             doit1(p2, p2) * doit2(p1, p2 - 1) + doit2(p2, p2) * doit1(p1, p2);
    } else {
      res += doit1(p1, p2) * doit2(q1, q2) + doit2(p1, p2) * doit1(q1, q2);
    }
  }
  printf("%.15lf", res);
  return 0;
}