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 main() {
  long long int pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  vector<long long int> v;
  vector<long long int> prev;
  v.push_back(4);
  v.push_back(7);
  prev.push_back(4);
  prev.push_back(7);
  for (int i = 2; i <= 9; i++) {
    vector<long long int> temp;
    for (int j = 0; j < prev.size(); j++) {
      v.push_back(10 * prev[j] + 4);
      v.push_back(10 * prev[j] + 7);
      temp.push_back(10 * prev[j] + 4);
      temp.push_back(10 * prev[j] + 7);
    }
    prev = temp;
  }
  long long int cnt = 0, p = 0, q = 1000000000;
  for (int i = 0; (i + k - 1) < v.size(); i++) {
    long long int x1, x2, x3, x4;
    if (i == 0) {
      x1 = 0, x2 = v[i], x3 = v[i + k - 1], x4 = v[i + k] - 1;
    } else if ((i + k - 1) == (v.size() - 1)) {
      x1 = v[i - 1] + 1, x2 = v[i], x3 = v[i + k - 1], x4 = q;
    } else {
      x1 = v[i - 1] + 1, x2 = v[i], x3 = v[i + k - 1], x4 = v[i + k] - 1;
    }
    long long int z1, z2, z3, z4;
    z1 = max(x1, pl);
    z2 = min(x2, pr);
    z3 = max(x3, vl);
    z4 = min(x4, vr);
    if ((z2 - z1 + 1) >= 1 && (z4 - z3 + 1) >= 1) {
      cnt += (z2 - z1 + 1) * (z4 - z3 + 1);
    }
    z1 = max(x1, vl);
    z2 = min(x2, vr);
    z3 = max(x3, pl);
    z4 = min(x4, pr);
    if ((z2 - z1 + 1) >= 1 && (z4 - z3 + 1) >= 1) {
      cnt += (z2 - z1 + 1) * (z4 - z3 + 1);
    }
  }
  long long int x = (pr - pl + 1) * (vr - vl + 1);
  if (k == 1) {
    for (int i = 0; i < v.size(); i++) {
      if (v[i] >= pl && v[i] <= pr && v[i] >= vl && v[i] <= vr) cnt--;
    }
  }
  long double z = (((long double)cnt) / x);
  printf("%0.12Lf", z);
  return 0;
}