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 main() {
  vector<int> lucky(1, 0);
  for (int i = 1; i <= 9; i++) {
    for (int j = 0; j < (1 << i); j++) {
      int b = 1, l = 0;
      for (int k = 0; k < i; k++) {
        if (j & (1 << k)) {
          l += 4 * b;
        } else {
          l += 7 * b;
        }
        b *= 10;
      }
      lucky.push_back(l);
    }
  }
  lucky.push_back(1000000001);
  sort(lucky.begin(), lucky.end());
  int pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  double d = 0.0;
  for (size_t i0 = 1; i0 + k < lucky.size(); i0++) {
    int prange = min(pr, lucky[i0]) - max(pl, lucky[i0 - 1] + 1) + 1;
    int vrange = min(vr, lucky[i0 + k] - 1) - max(vl, lucky[i0 + k - 1]) + 1;
    if (prange > 0 && vrange > 0) {
      d += 1.0 * prange * vrange;
    }
  }
  for (size_t i0 = 1; i0 + k < lucky.size(); i0++) {
    int vrange = min(vr, lucky[i0]) - max(vl, lucky[i0 - 1] + 1) + 1;
    int prange = min(pr, lucky[i0 + k] - 1) - max(pl, lucky[i0 + k - 1]) + 1;
    if (prange > 0 && vrange > 0) {
      d += 1.0 * prange * vrange;
    }
  }
  if (k == 1) {
    for (size_t i = 1; i < lucky.size() - 1; i++) {
      if (pl <= lucky[i] && lucky[i] <= pr && vl <= lucky[i] &&
          lucky[i] <= vr) {
        d -= 1.0;
      }
    }
  }
  printf("%.12f\n", d / (pr - pl + 1) / (vr - vl + 1));
  return 0;
}