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 plo, phi, vlo, vhi, k;
long long cnt = 0;
vector<int> lucky;
void make_lucky() {
  lucky.push_back(0);
  for (int i = 1; i <= 9; i++) {
    for (int j = 0; j < (1 << i); j++) {
      int c = j;
      int num = 0;
      int pow10 = 1;
      for (int k = 0; k < i; k++) {
        num += 4 * pow10;
        if (c % 2 == 1) {
          num += 3 * pow10;
        }
        c /= 2;
        pow10 *= 10;
      }
      lucky.push_back(num);
    }
  }
  lucky.push_back((long long)(1e9 + 1));
}
long long cover(int lo1, int hi1, int lo2, int hi2) {
  int lo = max(lo1, lo2);
  int hi = min(hi1, hi2);
  return max(0, hi - lo + 1);
}
int main() {
  cin >> plo >> phi >> vlo >> vhi >> k;
  make_lucky();
  for (int i = 1; i < lucky.size() - k; i++) {
    int lolo = lucky[i - 1] + 1;
    int lohi = lucky[i];
    int hilo = lucky[i + k - 1];
    int hihi = lucky[i + k] - 1;
    cnt += cover(lolo, lohi, plo, phi) * cover(hilo, hihi, vlo, vhi) +
           cover(lolo, lohi, vlo, vhi) * cover(hilo, hihi, plo, phi);
    if (k == 1 && plo <= lucky[i] && lucky[i] <= phi && vlo <= lucky[i] &&
        lucky[i] <= vhi)
      cnt--;
  }
  cout.precision(50);
  cout << 1.0 * cnt / (phi - plo + 1) / (vhi - vlo + 1);
}