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 MAX_VAL = 1e9;
int pl, pr, vl, vr, k;
string state = "";
vector<int> luckies;
void gen() {
  if (state.size() == 10) {
    return;
  }
  if (!state.empty()) {
    luckies.push_back(stoi(state));
  }
  state += '4';
  gen();
  state.pop_back();
  state += '7';
  gen();
  state.pop_back();
}
pair<int, int> intersection(int a, int b, int c, int d) {
  return make_pair(max(a, c), min(b, d));
}
int int_size(pair<int, int> i) { return max(0, i.second - i.first + 1); }
double calc(int a, int b, int c, int d) {
  double p = 0.0;
  pair<int, int> int_a = intersection(pl, pr, a, b);
  pair<int, int> int_b = intersection(vl, vr, c, d);
  p += ((double)int_size(int_a) / (double)(pr - pl + 1)) *
       ((double)int_size(int_b) / (vr - vl + 1));
  pair<int, int> int_c = intersection(pl, pr, c, d);
  pair<int, int> int_d = intersection(vl, vr, a, b);
  p += ((double)int_size(int_c) / (double)(pr - pl + 1)) *
       ((double)int_size(int_d) / (vr - vl + 1));
  return p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> pl >> pr >> vl >> vr >> k;
  gen();
  sort(luckies.begin(), luckies.end());
  double ans = 0.0;
  for (int i = 0; i + k - 1 < (int)luckies.size(); i++) {
    int j = i + k - 1;
    int l = (i == 0) ? 1 : luckies[i - 1] + 1;
    int r = (j == (int)luckies.size() - 1) ? MAX_VAL : luckies[j + 1] - 1;
    ans += calc(l, luckies[i], luckies[j], r);
    if (k == 1 && pl <= luckies[i] && luckies[i] <= pr && vl <= luckies[i] &&
        luckies[i] <= vr) {
      ans -= ((double)1 / (pr - pl + 1)) * ((double)1 / (vr - vl + 1));
    }
  }
  cout << fixed << setprecision(15) << ans << '\n';
  return 0;
}