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;
long long int powmod(long long int a, int b, int n) {
  long long int rm = 1;
  while (b) {
    if (b % 2) {
      rm = (rm * a) % n;
    }
    a = (a * a) % n;
    b /= 2;
  }
  return rm;
}
int k, lucky[int(1e5 + 10)], n = 0, a[10][500000], ct[10] = {0};
double ans = 0.0;
long long int p1, p2, v1, v2, l1, l2, r1, r2, i1, i2, i3, i4;
void brute(int level, int numbersofar) {
  if (level > 1) a[level - 1][++ct[level - 1]] = numbersofar;
  if (level == 10) {
    return;
  }
  brute(level + 1, numbersofar * 10 + 4);
  brute(level + 1, numbersofar * 10 + 7);
}
int main() {
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  double pair = (p2 - p1 + 1) * (v2 - v1 + 1);
  brute(1, 0);
  for (int i = 1; i <= 9; ++i) {
    for (int j = 1; j <= ct[i]; ++j) {
      lucky[++n] = a[i][j];
    }
  }
  lucky[0] = 0;
  for (int i = 1; i <= n - k + 1; ++i) {
    l1 = lucky[i - 1] + 1, l2 = lucky[i];
    if (i == n - k + 1) {
      r1 = lucky[i + k - 1], r2 = 1e9;
    } else {
      r1 = lucky[i + k - 1], r2 = lucky[i + k] - 1;
    }
    i1 = max(l1, v1);
    i2 = min(l2, v2);
    i3 = max(r1, p1);
    i4 = min(r2, p2);
    if (i2 >= i1 && i4 >= i3) {
      ans += ((i4 - i3 + 1) * (i2 - i1 + 1) * 1.0) / pair;
    }
    i1 = max(r1, v1);
    i2 = min(r2, v2);
    i3 = max(l1, p1);
    i4 = min(l2, p2);
    if (i2 >= i1 && i4 >= i3) {
      ans += ((i4 - i3 + 1) * (i2 - i1 + 1) * 1.0) / pair;
    }
    if (k == 1 && lucky[i] >= v1 && lucky[i] <= v2 && lucky[i] >= v1 &&
        lucky[i] <= v2)
      ans -= 1.0 / pair;
  }
  printf("%.10lf\n", ans);
  return 0;
}