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 f[2000];
int main() {
  int t = 0;
  f[t++] = 0;
  for (int i = 0; i < t; i++) {
    f[t++] = f[i] * 10 + 4;
    f[t++] = f[i] * 10 + 7;
    if (f[t - 1] > 1000000000) break;
  }
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  long long ans = 0;
  for (int l = 1, r = k; r < t; l++, r++) {
    int qq = 0;
    if (pl <= f[l] && vr >= f[r] && vl < f[r + 1] && pr > f[l - 1]) {
      qq = 1;
      ans += (min(f[l], pr) - max(pl, f[l - 1] + 1) + 1) *
             (min(vr, f[r + 1] - 1) - max(vl, f[r]) + 1);
    }
    if (vl <= f[l] && pr >= f[r] && pl < f[r + 1] && vr > f[l - 1]) {
      if (qq && k == 1) ans--;
      ans += (min(f[l], vr) - max(vl, f[l - 1] + 1) + 1) *
             (min(pr, f[r + 1] - 1) - max(pl, f[r]) + 1);
    }
  }
  long long sum = (pr - pl + 1) * (vr - vl + 1);
  printf("%.12lf\n", (double)ans / sum);
  return 0;
}