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 long long N = 1e5 + 5;
long long l1, r1, l2, r2, k;
vector<long long> v;
void rec(long long x) {
  if (x <= 1e9) v.push_back(x);
  if (x > 1e9) return;
  rec(x * 10 + 4);
  rec(x * 10 + 7);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  rec(0);
  v.push_back(2e9);
  sort(v.begin(), v.end());
  long long n = v.size();
  double num = 0;
  double den = (r2 - l2 + 1) * (r1 - l1 + 1);
  for (long long i = 0; i + k + 1 < n; i++) {
    long long j = i + k;
    long long range1 = min(v[i + 1], r1) - max(v[i] + 1, l1) + 1;
    long long range2 = min(v[j + 1] - 1, r2) - max(v[j], l2) + 1;
    if (range1 >= 0 && range2 >= 0) num += range1 * range2;
  }
  for (long long i = k; i < n; i++) {
    long long j = i - k;
    long long range1 = min(v[i + 1] - 1, r1) - max(v[i], l1) + 1;
    long long range2 = min(v[j + 1], r2) - max(v[j] + 1, l2) + 1;
    if (range1 >= 0 && range2 >= 0) num += range1 * range2;
  }
  if (k == 1) {
    for (auto &it : v) {
      if (it >= l1 && it <= r1 && it >= l2 && it <= r2) num--;
    }
  }
  double ans = num / den;
  cout << fixed << setprecision(12) << ans;
  return 0;
}