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 gera(long long at) {
  if (at > 1000000000ll) return;
  if (at) luck.push_back((int)at);
  gera(10ll * at + 4);
  gera(10ll * at + 7);
}
double prob(int a, int b, int ini, int fim) {
  double tam = b - a + 1;
  double inter = min(b, fim) - max(a, ini) + 1;
  if (inter < 0) inter = 0;
  return inter / tam;
}
int main() {
  luck.clear();
  gera(0);
  luck.push_back(0);
  luck.push_back(1000000001);
  sort(luck.begin(), luck.end());
  int a, b, c, d, k;
  cin >> a >> b >> c >> d >> k;
  double ans = 0;
  int ini = 1, fim = k;
  while (fim + 1 < (int)(luck).size()) {
    ans += prob(a, b, luck[ini - 1] + 1, luck[ini]) *
           prob(c, d, luck[fim], luck[fim + 1] - 1);
    ans += prob(c, d, luck[ini - 1] + 1, luck[ini]) *
           prob(a, b, luck[fim], luck[fim + 1] - 1);
    if (k == 1)
      ans -=
          prob(c, d, luck[ini], luck[ini]) * prob(a, b, luck[fim], luck[fim]);
    fim++;
    ini++;
  }
  cout.precision(15);
  cout << fixed << ans << endl;
  return 0;
}