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;
unsigned X[1024], Y[1024];
int main(void) {
  X[0] = 0;
  X[1] = 4;
  X[2] = 7;
  int s = 3;
  for (int i = 2, j = 10; i <= 9; i++, j *= 10) {
    for (int k = 0; k < 1 << (i - 1); k++) {
      X[s++] = j * 4 + X[(1 << (i - 1)) - 1 + k];
    }
    for (int k = 0; k < 1 << (i - 1); k++) {
      X[s++] = j * 7 + X[(1 << (i - 1)) - 1 + k];
    }
  }
  X[1023] = 1000000001;
  for (int i = 0; i < 1024; i++) {
    Y[i] = X[i];
  }
  unsigned ai, bi, ci, di, a, b, c, d, k;
  cin >> a >> b >> c >> d >> k;
  for (ai = 0; a > X[ai]; ai++) {
  }
  X[--ai] = a;
  for (bi = 1023; b < X[bi]; bi--) {
  }
  X[++bi] = b;
  for (ci = 0; c > Y[ci]; ci++) {
  }
  Y[--ci] = c;
  for (di = 1023; d < Y[di]; di--) {
  }
  Y[++di] = d;
  unsigned long long res = 0, all;
  for (unsigned i = ai; i < bi; i++)
    for (unsigned j = ci; j < di; j++)
      if (i - j == k || j - i == k) {
        res += (unsigned long long)(X[i + 1] - X[i] +
                                    ((i < j && i == ai || i > j && i == bi - 1)
                                         ? 1
                                         : 0)) *
               (Y[j + 1] - Y[j] +
                ((j < i && j == ci || j > i && j == di - 1) ? 1 : 0));
      }
  if (k == 1) {
    int r = min((int)bi, (int)di) - max((int)ai, (int)ci) - 1;
    if (r > 0) res -= r;
  }
  all = (unsigned long long)(b - a + 1) * (d - c + 1);
  cout.precision(15);
  cout << fixed << (double)res / all;
  return 0;
}