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;
vector<long long int> v;
void dfs(long long int n) {
  v.push_back(n);
  if (n * 10LL + 4LL < 1000000000LL) dfs(n * 10LL + 4LL);
  if (n * 10LL + 7LL < 1000000000LL) dfs(n * 10LL + 7LL);
}
long long int f(pair<long long int, long long int> rango, long long int a,
                long long int b) {
  if (rango.second < a || b < rango.first) return 0LL;
  if (rango.first <= a && b <= rango.second) return b - a + 1LL;
  if (a <= rango.first && rango.second <= b)
    return rango.second - rango.first + 1LL;
  if (rango.first >= a) return b - rango.first + 1LL;
  return rango.second - a + 1LL;
}
int main() {
  dfs(4);
  dfs(7);
  v.push_back(0);
  v.push_back(1000000000LL + 1);
  sort(v.begin(), v.end());
  pair<long long int, long long int> a, b;
  long long int cnt, sum = 0LL;
  cin >> a.first >> a.second >> b.first >> b.second >> cnt;
  for (int i = 0; i + cnt < v.size(); i++) {
    if (i == 0 || i == v.size() - 1) continue;
    long long int tmpA = f(a, v[i - 1] + 1LL, v[i]);
    long long int tmpB = f(b, v[i + cnt - 1], v[i + cnt] - 1LL);
    sum += tmpA * tmpB;
    tmpA = f(b, v[i - 1] + 1LL, v[i]);
    tmpB = f(a, v[i + cnt - 1], v[i + cnt] - 1LL);
    sum += tmpA * tmpB;
  }
  long long int lenA = a.second - a.first + 1LL;
  long long int lenB = b.second - b.first + 1LL;
  if (cnt == 1) {
    for (int i = 1; i < v.size() - 1; i++) {
      if (a.first <= v[i] && v[i] <= a.second && b.first <= v[i] &&
          v[i] <= b.second)
        sum--;
    }
  }
  printf("%.12lf\n", (double)sum / (lenA * lenB));
  return 0;
}