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;
const int Max = 1000000100;
vector<long long> s;
void dfs(long long temp) {
  if (temp > Max) {
    return;
  }
  long long x = temp * 10 + 4;
  if (temp > Max) {
    return;
  }
  s.push_back(x);
  dfs(x);
  long long y = temp * 10 + 7;
  if (temp > Max) {
    return;
  }
  s.push_back(y);
  dfs(y);
}
int main() {
  dfs(0);
  s.push_back(0);
  sort(s.begin(), s.end());
  long long l, r, x, y;
  int n;
  cin >> l >> r >> x >> y >> n;
  long long temp = l > x ? x : l;
  int i = lower_bound(s.begin(), s.end(), temp) - s.begin();
  double sum = 0;
  while (true) {
    long long a = s[i - 1] + 1;
    long long b = s[i];
    long long c = s[i + n - 1];
    long long d = s[i + n] - 1;
    if (a > r && a > y) break;
    if (c > r && c > y) break;
    long long A = a > l ? a : l;
    long long B = b < r ? b : r;
    long long C = c > x ? c : x;
    long long D = d < y ? d : y;
    int shabi;
    int z = 0;
    if (n == 1 && B == C) {
      int o = lower_bound(s.begin(), s.end(), B) - s.begin();
      if (s[o] == B) z = 1;
    }
    if (z) shabi = D;
    if (A <= B && C <= D) {
      if (z)
        sum += (double)(B - A) / (double)(r - l + 1) * (double)(D - C + 1) /
               (double)(y - x + 1);
      else
        sum += (double)(B - A + 1) / (double)(r - l + 1) * (double)(D - C + 1) /
               (double)(y - x + 1);
    }
    A = a > x ? a : x;
    B = b < y ? b : y;
    C = c > l ? c : l;
    D = d < r ? d : r;
    if (A <= B && C <= D) {
      if (z)
        sum += (double)(B - A + 1) / (double)(y - x + 1) * (double)(D - C) /
               (double)(r - l + 1);
      else
        sum += (double)(B - A + 1) / (double)(y - x + 1) * (double)(D - C + 1) /
               (double)(r - l + 1);
    }
    if (z) {
      sum += (double)(shabi - A + 1) / (double)(y - x + 1) * (double)(1) /
             (double)(r - l + 1);
    }
    i++;
  }
  printf("%.12lf", sum);
  return 0;
}