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;
vector<int> luck;
void lucky(long long v) {
  if (v > 1e9) return;
  luck.push_back(v);
  lucky(v * 10 + 4);
  lucky(v * 10 + 7);
}
long long intersect(int a, int b, int x, int y) {
  a = max(a, x);
  b = min(b, y);
  return max(0ll, 1ll * (b - a + 1));
}
pair<int, int> inter(int a, int b, int x, int y) {
  a = max(a, x);
  b = min(b, y);
  if (a <= b) return make_pair(a, b);
  return make_pair(0, -1);
}
int main() {
  lucky(4);
  lucky(7);
  int a, b, x, y, k;
  scanf("%d %d %d %d %d", &a, &b, &x, &y, &k);
  luck.push_back(0);
  luck.push_back(1e9 + 5);
  sort(luck.begin(), luck.end());
  long long t = 0;
  for (int i = 1; i + k < (int)luck.size(); ++i) {
    t += intersect(a, b, luck[i - 1] + 1, luck[i]) *
         intersect(x, y, luck[i + k - 1], luck[i + k] - 1);
    t += intersect(x, y, luck[i - 1] + 1, luck[i]) *
         intersect(a, b, luck[i + k - 1], luck[i + k] - 1);
    pair<int, int> p1 = inter(a, b, luck[i - 1] + 1, luck[i]);
    pair<int, int> p2 = inter(x, y, luck[i + k - 1], luck[i + k] - 1);
    pair<int, int> p3 = inter(x, y, luck[i - 1] + 1, luck[i]);
    pair<int, int> p4 = inter(a, b, luck[i + k - 1], luck[i + k] - 1);
    p1 = inter(p1.first, p1.second, p2.first, p2.second);
    p3 = inter(p3.first, p3.second, p4.first, p4.second);
    p1 = inter(p1.first, p1.second, p3.first, p3.second);
    t -= (1ll * p1.second - p1.first + 1) * (1ll * p1.second - p1.first + 1);
  }
  printf("%.15lf", t * 1.0 / (1ll * (y - x + 1) * (b - a + 1)));
}