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;
template <class T>
T gcd(T a, T b) {
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T>
T lcm(T a, T b) {
  return a * (b / gcd(a, b));
}
using namespace std;
vector<long long> L;
map<long long, bool> mapa;
void go(long long now) {
  if (now > 1000000000LL) return;
  if (now > 0 && mapa[now] == 0) {
    L.push_back(now);
  }
  go(now * 10LL + 4LL);
  go(now * 10LL + 7LL);
}
int main() {
  go(0);
  sort(L.begin(), L.end());
  int len = L.size();
  long long p1, pr, v1, vr, k;
  while (cin >> p1 >> pr >> v1 >> vr >> k) {
    long long tot = 0;
    for (int i = 0; i < len - k + 1; i++) {
      long long awal = L[i];
      long long akhir = L[i + k - 1];
      long long temp = 0;
      if (p1 <= awal && akhir <= vr) {
        if (i)
          temp = min(awal, pr) - max(p1, L[i - 1] + 1LL) + 1LL;
        else
          temp = min(awal, pr) - p1 + 1LL;
        if (i + k < len && temp > 0)
          temp *= (min(vr, L[i + k] - 1LL) - max(v1, L[i + k - 1]) + 1LL);
        else
          temp *= (vr - max(v1, L[i + k - 1]) + 1LL);
      }
      swap(p1, v1);
      swap(pr, vr);
      if (temp > 0) tot += temp;
      temp = 0;
      if (p1 <= awal && akhir <= vr) {
        if (i)
          temp = min(awal, pr) - max(p1, L[i - 1] + 1LL) + 1LL;
        else
          temp = min(awal, pr) - p1 + 1LL;
        if (i + k < len && temp > 0)
          temp *= (min(vr, L[i + k] - 1LL) - max(v1, L[i + k - 1]) + 1LL);
        else
          temp *= (vr - max(v1, L[i + k - 1]) + 1LL);
        if (temp > 0 && k == 1) temp--;
      }
      if (temp > 0) tot += temp;
      swap(p1, v1);
      swap(pr, vr);
    }
    double ans =
        (double)tot / (double)(vr - v1 + 1LL) / (double)(pr - p1 + 1LL);
    printf("%.9lf\n", ans);
  }
  return 0;
}