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<long long> lucky = {4, 7};
  map<long long, long long> next, prev;
  next[4] = 7;
  prev[4] = 0;
  for (int i = 0; i < 1025; ++i) {
    next[lucky[i]] = lucky[i + 1] - 1;
    if (i > 0) {
      prev[lucky[i]] = lucky[i - 1] + 1;
    }
    lucky.push_back(lucky[i] * 10 + 4);
    lucky.push_back(lucky[i] * 10 + 7);
  }
  map<pair<long long, long long>, long long> count;
  for (int i = 0; i < lucky.size(); ++i) {
    for (int j = i; j < lucky.size(); ++j) {
      count[{lucky[i], lucky[j]}] = j - i + 1;
    }
  }
  long long l1, r1, l2, r2, k;
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  long long answer = 0;
  for (auto i : count) {
    if (i.second == k) {
      answer +=
          max(max((min(r1, i.first.first) - max(l1, prev[i.first.first]) + 1),
                  (long long)0) *
                  max(((min(r2, next[i.first.second]) -
                        max(l2, i.first.second) + 1)),
                      (long long)0),
              (long long)0);
      answer +=
          max(max((min(r2, i.first.first) - max(l2, prev[i.first.first]) + 1),
                  (long long)0) *
                  max(((min(r1, next[i.first.second]) -
                        max(l1, i.first.second) + 1)),
                      (long long)0),
              (long long)0);
      if (k == 1 and (i.first.first >= l1 and i.first.first <= r1 and
                      i.first.first >= l2 and i.first.first <= r2)) {
        answer--;
      }
    }
  }
  cout << fixed << setprecision(10)
       << (long double)answer / ((r2 - l2 + 1) * (r1 - l1 + 1));
  return 0;
}