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 INF = 1000000000;
template <class T>
inline int size(const T& c) {
  return c.size();
}
inline int getint() {
  int x;
  scanf("%d", &x);
  return x;
}
vector<long long> lucky;
long long A, B;
int calc(int a, int b) {
  int ret = 0;
  for (__typeof(lucky.begin()) it = lucky.begin(); it != lucky.end(); ++it)
    if (*it >= a && *it <= b) ret++;
  return ret;
}
long long brutal(int p1, int p2, int v1, int v2, int k) {
  long long ret = 0;
  for (int a = (p1), _b = (p2); a <= _b; ++a)
    for (int b = (v1), _b = (v2); b <= _b; ++b)
      if (calc(min(a, b), max(a, b)) == k) {
        ret++;
        cerr << "ret"
             << " = " << ret << ", "
             << "a"
             << " = " << a << ", "
             << "b"
             << " = " << b << endl;
        ;
      }
  return ret;
}
void genLucky(long long x) {
  if (x > B) return;
  if (x >= A && x <= B) lucky.push_back(x);
  genLucky(x * 10LL + 4);
  genLucky(x * 10LL + 7);
}
inline long long cross(long long a1, long long a2, long long b1, long long b2) {
  return max(0LL, min(a2, b2) - max(a1, b1) + 1);
}
double calc(long long p1, long long p2, long long v1, long long v2, int k) {
  long long atas = 0;
  long long bawah = (p2 - p1 + 1) * (v2 - v1 + 1);
  for (int from = 0, _n = (size(lucky) - k + 1); from < _n; ++from) {
    int to = from + k - 1;
    long long alfa = lucky[from];
    long long prevAlfa = from == 0 ? 0 : lucky[from - 1];
    long long beta = lucky[to];
    long long nextBeta = to == size(lucky) - 1 ? (INF + 1) : lucky[to + 1];
    prevAlfa++;
    nextBeta--;
    atas += cross(p1, p2, prevAlfa, alfa) * cross(v1, v2, beta, nextBeta);
    atas += cross(v1, v2, prevAlfa, alfa) * cross(p1, p2, beta, nextBeta);
    if (k == 1) atas -= cross(v1, v2, alfa, alfa) * cross(p1, p2, alfa, alfa);
  }
  return 1.0 * atas / bawah;
}
int main() {
  int p1 = getint();
  int p2 = getint();
  int v1 = getint();
  int v2 = getint();
  int k = getint();
  A = min(p1, v1);
  B = max(p2, v2);
  genLucky(0);
  sort(lucky.begin(), lucky.end());
  printf("%.10lf\n", calc(p1, p2, v1, v2, k));
  return 0;
}