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 k;
vector<int> v;
bool isluk(int x) {
  while (x) {
    int d = x % 10;
    if (d != 4 && d != 7) return false;
    x /= 10;
  }
  return true;
}
int getsz(int a, int b, int x, int y) {
  int s = max(x, a);
  int e = min(b, y);
  return max(0, e - s + 1);
}
double calc(int pl, int pr, int vl, int vr) {
  long long sum = 0;
  for (int i = 0; i + k + 1 < v.size(); i++) {
    int f = v[i], s = v[i + k];
    long long t1 = (double)getsz(v[i] + 1, v[i + 1], pl, pr);
    long long t2 = (double)getsz(v[i + k], v[i + k + 1] - 1, vl, vr);
    sum += t1 * t2;
  }
  double ret = (double)sum / (vr - vl + 1) / (pr - pl + 1);
  return ret;
}
int main() {
  int pl, pr, vl, vr;
  while (cin >> pl >> pr >> vl >> vr >> k) {
    v.clear();
    for (int i = 1; i < 10; i++) {
      for (int j = 0; j < (1 << i); j++) {
        int x = 0;
        for (int k = 0; k < i; k++) x = 10 * x + ((j & (1 << k)) ? 4 : 7);
        v.push_back(x);
      }
    }
    v.push_back(0);
    v.push_back(1000000001);
    sort(v.begin(), v.end());
    double res = 0;
    res += calc(pl, pr, vl, vr);
    res += calc(vl, vr, pl, pr);
    if (k == 1) {
      int cnt = 0;
      for (int i = 1; i + 1 < v.size(); i++)
        if (v[i] >= pl && v[i] <= pr && v[i] >= vl && v[i] <= vr) cnt++;
      res -= (double)cnt / (vr - vl + 1) / (pr - pl + 1);
    }
    cout.precision(10);
    cout.setf(ios::fixed | ios::showpoint);
    cout << res << endl;
  }
  return 0;
}