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;
const long long MAX_N = 1000000000LL;
vector<int> lucky;
void find(long long n) {
  if (n > MAX_N) return;
  lucky.push_back((int)n);
  find(n * 10LL + 4LL);
  find(n * 10LL + 7LL);
}
int main() {
  int pl, pr, vl, vr, k;
  long long good = 0;
  long long left, right;
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k);
  find(4);
  find(7);
  sort(lucky.begin(), lucky.end());
  lucky.push_back(2000000000);
  for (int i = 0; i < lucky.size() - k + 1; ++i) {
    if (lucky[i] >= pl && (i == 0 || pr >= lucky[i - 1]) &&
        lucky[i + k - 1] <= vr && vl < lucky[i + k]) {
      if (i == 0)
        left = lucky[i] - pl + 1;
      else
        left = min(lucky[i], pr) - max(pl - 1, lucky[i - 1]);
      if (i == lucky.size() - k + 1)
        right = vr - lucky[i + k - 1] + 1;
      else
        right = min(lucky[i + k], vr + 1) - max(lucky[i + k - 1], vl);
      good += left * right;
    }
    if (lucky[i] >= vl && (i == 0 || vr >= lucky[i - 1]) &&
        lucky[i + k - 1] <= pr && pl < lucky[i + k]) {
      if (i == 0)
        left = lucky[i] - vl + 1;
      else
        left = min(lucky[i], vr) - max(vl - 1, lucky[i - 1]);
      if (i == lucky.size() - k + 1)
        right = pr - lucky[i + k - 1] + 1;
      else
        right = min(lucky[i + k], pr + 1) - max(lucky[i + k - 1], pl);
      good += left * right;
      if (lucky[i] == lucky[i + k - 1]) --good;
    }
  }
  double ret = (double)good / ((double)(pr - pl + 1) * (double)(vr - vl + 1));
  printf("%.12lf\n", ret);
  return 0;
}