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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int Mod = 998244353;
const long double Pi = acos(-1);
const long long int Inf = 4e18;
const long double Eps = 1e-9;
int dx[9] = {0, 1, -1, 0, 0, 1, 1, -1, -1};
int dy[9] = {0, 0, 0, 1, -1, 1, -1, 1, -1};
using namespace std;
long long int Intersection(long long int l1, long long int r1, long long int l2,
                           long long int r2) {
  return max(0LL, min(r1, r2) - max(l1, l2) + 1);
}
void TestCase() {
  long long int pl, pr, vl, vr, k, tot = 0;
  cin >> pl >> pr >> vl >> vr >> k;
  long double ans;
  vector<long long int> num;
  for (long long int len = 1; len <= 9; len++) {
    for (long long int i = 0; i < (1 << len); i++) {
      long long int p = 1, n = 0;
      for (long long int j = 0; j < len; j++) {
        if (i & (1 << j))
          n += 7 * p;
        else
          n += 4 * p;
        p = p * 10;
      }
      num.push_back(n);
    }
  }
  sort(num.begin(), num.end());
  num.insert(num.begin(), 0);
  for (long long int i = 1; i < (long long int)num.size(); i++) {
    if (i + k > (long long int)num.size()) continue;
    long long int a, b, c, d, up;
    if (i + k == (long long int)num.size())
      up = 1000000000;
    else
      up = num[i + k] - 1;
    a = Intersection(pl, pr, num[i - 1] + 1, num[i]);
    b = Intersection(vl, vr, num[i + k - 1], up);
    c = Intersection(vl, vr, num[i - 1] + 1, num[i]);
    d = Intersection(pl, pr, num[i + k - 1], up);
    tot += a * b + c * d;
    if (k == 1) {
      bool ok =
          ((num[i] >= pl && num[i] <= pr) && (num[i] >= vl && num[i] <= vr));
      if (ok) tot--;
    }
  }
  long double f = (long double)(tot) / ((vr - vl + 1) * (pr - pl + 1));
  cout << fixed << setprecision(10) << f;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int T = 1;
  while (T--) {
    TestCase();
    cout << "\n";
  }
  return 0;
}