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> num;
const long long MAX = (long long)1E10;
void rec(long long x) {
  if (x > MAX) return;
  num.push_back(x * 10 + 4);
  num.push_back(x * 10 + 7);
  rec(x * 10 + 4);
  rec(x * 10 + 7);
}
long long pl, pr, vl, vr;
int k;
long double prob_p(long long a, long long b) {
  if (a > b || a > pr || b < pl) return 0;
  return (long double)(min(pr, b) - max(pl, a) + 1) / (pr - pl + 1);
}
long double prob_v(long long a, long long b) {
  if (a > b || a > vr || b < vl) return 0;
  return (long double)(min(vr, b) - max(vl, a) + 1) / (vr - vl + 1);
}
int main() {
  rec(0);
  num.push_back(-MAX);
  num.push_back(MAX);
  sort(num.begin(), num.end());
  cin >> pl >> pr >> vl >> vr >> k;
  long double ans = 0;
  int n = num.size();
  if (k > 1) {
    for (int i = 1; i + 2 < n; ++i) {
      long long a = num[i - 1] + 1, b = num[i];
      long long c = num[i + k - 1], d = num[i + k] - 1;
      ans += prob_p(a, b) * prob_v(c, d) + prob_v(a, b) * prob_p(c, d);
    }
  } else {
    for (int i = 1; i + 1 < n; ++i) {
      long long a = num[i - 1] + 1, b = num[i + 1] - 1;
      long long x = num[i];
      ans += prob_p(a, x - 1) * prob_v(x + 1, b) +
             prob_v(a, x - 1) * prob_p(x + 1, b) +
             +prob_p(a, x - 1) * prob_v(x, x) +
             prob_v(a, x - 1) * prob_p(x, x) +
             +prob_p(x + 1, b) * prob_v(x, x) +
             prob_v(x + 1, b) * prob_p(x, x) + prob_p(x, x) * prob_v(x, x);
    }
  }
  cout << fixed << setprecision(12) << ans << '\n';
  return 0;
}