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 PL, PR, VL, VR, K;
vector<int> s;
int main() {
  long long ret = 0;
  scanf("%d %d %d %d %d", &PL, &PR, &VL, &VR, &K);
  for (int i = 1; i <= 9; i++) {
    for (int j = 0; j < (1 << i); j++) {
      int x = 0;
      for (int k = 0; k < i; k++) {
        if ((bool)(j & (1 << k)) == 0) {
          x *= 10;
          x += 4;
        } else {
          x *= 10;
          x += 7;
        }
      }
      s.push_back(x);
    }
  }
  sort(s.begin(), s.end());
  int N = s.size();
  for (int i = 0; i + K <= N; i++) {
    int pl = 1, pr = s[i], vl = s[i + K - 1], vr = 1000000000, f = 0;
    if (i) pl = s[i - 1] + 1;
    if (i + K < N) vr = s[i + K] - 1;
    pl = max(pl, PL);
    pr = min(pr, PR);
    vl = max(vl, VL);
    vr = min(vr, VR);
    if (pr >= pl && vr >= vl)
      ret += (long long)(pr - pl + 1) * (long long)(vr - vl + 1), f++;
    pl = 1, pr = s[i], vl = s[i + K - 1], vr = 1000000000;
    if (i) pl = s[i - 1] + 1;
    if (i + K < N) vr = s[i + K] - 1;
    pl = max(pl, VL);
    pr = min(pr, VR);
    vl = max(vl, PL);
    vr = min(vr, PR);
    if (pr >= pl && vr >= vl)
      ret += (long long)(pr - pl + 1) * (long long)(vr - vl + 1), f++;
    if (f == 2 && K == 1) ret--;
  }
  printf("%.9lf\n",
         (double)ret / ((double)(PR - PL + 1) * (double)(VR - VL + 1)));
}