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 N = 100000, MOD = 1e9 + 7;
const int OO = 1e9 + 5000;
vector<long long> sol;
long long firstleft, firstright, secondleft, secondright, k;
void generate(int i, long long number) {
  sol.push_back(number);
  if (number > (int)1e9) {
    return;
  }
  generate(i + 1, number * 10 + 4), generate(i + 1, number * 10 + 7);
}
long long getFirst(long long from, long long to) {
  from = max(from, firstleft);
  to = min(to, firstright);
  return max(0ll, to - from + 1);
}
long long getSecond(long long from, long long to) {
  from = max(from, secondleft);
  to = min(to, secondright);
  return max(0ll, to - from + 1);
}
int main() {
  generate(0, 0);
  sort(sol.begin(), sol.end());
  int tc;
  tc = 1;
  while (tc-- > 0) {
    cin >> firstleft >> firstright >> secondleft >> secondright >> k;
    long long ans = 0;
    for (int i = 1; i < 2047; ++i) {
      if (i + k >= 2047) break;
      long long c1 = getFirst(sol[i - 1] + 1, sol[i]) *
                     getSecond(sol[i + k - 1], sol[i + k] - 1);
      long long c2 = getFirst(sol[i + k - 1], sol[i + k] - 1) *
                     getSecond(sol[i - 1] + 1, sol[i]);
      long long toadd = (c1 + c2);
      if (k == 1 && sol[i] >= firstleft && sol[i] <= firstright &&
          sol[i] >= secondleft && sol[i] <= secondright) {
        --toadd;
      }
      ans += toadd;
    }
    double val = ans;
    double range = (firstright - firstleft + 1);
    range *= double(secondright - secondleft + 1);
    cout << fixed << setprecision(10) << ans / range << endl;
  }
}