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 m = 1000000000;
std::vector<int> L;
int mi, ma;
void findLuckyNr(int n) {
  if (n > ma) return;
  if (n >= mi) L.push_back(n);
  if (n > ma / 10) return;
  findLuckyNr(n * 10 + 4);
  findLuckyNr(n * 10 + 7);
}
int main(void) {
  int i, j;
  int cop, cov, k, pl, pr, vl, vr, last_l, next_r;
  double ans = 0.0, den = 0.0;
  int l00, l01, r00, r01;
  int l10, l11, r10, r11;
  cin >> pl >> pr >> vl >> vr >> k;
  mi = ((pl) < (vl) ? (pl) : (vl)), ma = ((pr) >= (vr) ? (pr) : (vr));
  findLuckyNr(4);
  findLuckyNr(7);
  sort(L.begin(), L.end());
  den = (double)(pr - pl + 1) * (double)(vr - vl + 1);
  for ((i) = 0; (i) < (int)(L.size() - k + 1); (i)++) {
    if (i) last_l = L[i - 1] + 1;
    if (i < L.size() - k) next_r = L[i + k] - 1;
    if (!i) last_l = pl;
    if (i == L.size() - k) next_r = vr;
    l00 = l01 = l10 = l11 = r00 = r01 = r10 = r11 = -1;
    if (pl <= L[i] && pr >= last_l && vr >= L[i + k - 1] && vl <= next_r) {
      l01 = ((L[i]) < (pr) ? (L[i]) : (pr));
      l00 = ((last_l) >= (pl) ? (last_l) : (pl));
      r01 = ((next_r) < (vr) ? (next_r) : (vr));
      r00 = ((vl) >= (L[i + k - 1]) ? (vl) : (L[i + k - 1]));
      cop = l01 - l00 + 1;
      cov = r01 - r00 + 1;
      ans += (double)cop * (double)cov;
    }
    if (!i) last_l = vl;
    if (i == L.size() - k) next_r = pr;
    if (vl <= L[i] && vr >= last_l && pr >= L[i + k - 1] && pl <= next_r) {
      l11 = ((L[i]) < (vr) ? (L[i]) : (vr));
      l10 = ((last_l) >= (vl) ? (last_l) : (vl));
      r11 = ((next_r) < (pr) ? (next_r) : (pr));
      r10 = ((pl) >= (L[i + k - 1]) ? (pl) : (L[i + k - 1]));
      cov = l11 - l10 + 1;
      cop = r11 - r10 + 1;
      ans += (double)cop * (double)cov;
      if (l01 == r00 && r00 == L[i] && l11 == r10 && r10 == L[i]) ans--;
    }
    if (pr <= L[i] || vr <= L[i]) break;
  }
  ans /= den;
  printf("%.9f\n", ans);
  return 0;
}