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 int SIZE = 50000;
int find_min(int a, int b, int mini, int bef) {
  if (mini < a || bef >= b) return 0;
  a = max(a, bef + 1);
  return min(mini, b) - a + 1;
}
int find_max(int a, int b, int maxi, int next) {
  if (maxi > b || next <= a) return 0;
  b = min(next - 1, b);
  return b - max(a, maxi) + 1;
}
int main() {
  int a, b, c, d, k;
  vector<int> lucky;
  queue<pair<int, int> > Q;
  Q.push(make_pair(0, 1));
  while (!Q.empty()) {
    pair<int, int> b = Q.front();
    Q.pop();
    if (b.first != 0) lucky.push_back(b.first);
    if (b.second == 1000000000) continue;
    Q.push(make_pair(b.first + (4 * b.second), b.second * 10));
    Q.push(make_pair(b.first + (7 * b.second), b.second * 10));
  }
  sort(lucky.begin(), lucky.end());
  while (cin >> a >> b >> c >> d >> k) {
    double ans = 0;
    k--;
    for (int i = 0; i < lucky.size(); i++) {
      int s = i + k, bef = -1, next = 1000000000;
      if (s >= lucky.size() || lucky[s] > max(b, d)) break;
      if (i != 0) bef = lucky[i - 1];
      if (s + 1 != lucky.size()) next = lucky[s + 1];
      double x = find_min(a, b, lucky[i], bef),
             y = find_min(c, d, lucky[i], bef);
      double w = find_max(a, b, lucky[s], next),
             h = find_max(c, d, lucky[s], next);
      double e = (x / (b - a + 1)) * (h / (d - c + 1)) +
                 (w / (b - a + 1)) * (y / (d - c + 1));
      if (k == 0) {
        if (a <= lucky[i] && lucky[i] <= b) {
          if (c <= lucky[i] && lucky[i] <= d) {
            e -= (1.0 / (b - a + 1) * (1.0 / (d - c + 1)));
          }
        }
      }
      if (e == 2.0) e--;
      ans += e;
    }
    printf("%.9f\n", ans);
  }
  return 0;
}