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 = 1e6 + 5;
vector<long long int> a(MAX);
long long int c;
void gen(long long int x) {
  if (x > 1e10) return;
  a[c++] = x;
  gen(x * 10 + 4);
  gen(x * 10 + 7);
}
long long int calc(long long int Pl, long long int Pr, long long int Vl,
                   long long int Vr) {
  return max(min(Pr, Vr) - max(Pl, Vl) + 1, 0LL);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  gen(0);
  sort(a.begin(), a.begin() + c);
  long long int Pl, Pr, Vl, Vr, k;
  cin >> Pl >> Pr >> Vl >> Vr >> k;
  long long int res = 0;
  for (long long int i = 1, j; i + k < c; i++) {
    j = i + k;
    long long int x = calc(a[i - 1] + 1, a[i], Pl, Pr);
    long long int y = calc(a[j - 1], a[j] - 1, Vl, Vr);
    res += x * y;
    x = calc(a[i - 1] + 1, a[i], Vl, Vr);
    y = calc(a[j - 1], a[j] - 1, Pl, Pr);
    res += x * y;
    if (k == 1 && Pl <= a[i] && a[i] <= Pr && Vl <= a[i] && a[i] <= Vr) res--;
  }
  double ans = ((1.0 * res) / (Pr - Pl + 1)) / (Vr - Vl + 1);
  cout << fixed << setprecision(12) << ans << '\n';
}