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 testnum;
vector<long long> all;
void preprocess(long long num) {
  if (num <= 1000000000ll) {
    if (num > 0) all.push_back(num);
    preprocess(num * 10 + 4);
    preprocess(num * 10 + 7);
  }
}
long long al, ah, bl, bh, k;
void solve() {
  int N = all.size();
  double sum = 0;
  for (int i = 0; i + k <= N; i++) {
    long long x = all[i];
    long long y = all[i + k - 1];
    long long px, py;
    if (al <= x && bh >= y) {
      if (i == 0)
        px = al;
      else
        px = max(al, all[i - 1] + 1);
      if (i + k == N)
        py = bh;
      else
        py = min(bh, all[i + k] - 1);
      double t1 = min(x, ah) - px + 1;
      double t2 = py - max(y, bl) + 1;
      if (t1 < 0) t1 = 0;
      if (t2 < 0) t2 = 0;
      sum += t1 * t2;
    }
    if (bl <= x && ah >= y) {
      if (i == 0)
        py = bl;
      else
        py = max(bl, all[i - 1] + 1);
      if (i + k == N)
        px = ah;
      else
        px = min(ah, all[i + k] - 1);
      double t1 = px - max(y, al) + 1;
      double t2 = min(x, bh) - py + 1;
      if (t1 < 0) t1 = 0;
      if (t2 < 0) t2 = 0;
      sum += t1 * t2;
    }
    if (k == 1 && al <= x && x <= ah && bl <= x && x <= bh) sum -= 1.0;
  }
  double tot = (double)(ah - al + 1) * (double)(bh - bl + 1);
  printf("%.12lf\n", sum / tot);
}
bool input() {
  cin >> al >> ah >> bl >> bh >> k;
  return true;
}
int main() {
  preprocess(0);
  sort(all.begin(), all.end());
  int T = 1;
  for (testnum = 1; testnum <= T; testnum++) {
    if (!input()) break;
    solve();
  }
}