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;
long long n1, n2, m1, m2, left, right, k;
int MAXV = 1000000000;
int n = 0;
vector<int> a;
void generate(long long x) {
  if (x <= MAXV) {
    if (x != 0) a.push_back(x);
  } else {
    return;
  }
  x = x * 10 + 4;
  generate(x);
  x = x - 4 + 7;
  generate(x);
}
long long g(long long l1, long long l2, long long n, long long m) {
  long long minx = max(l1, n), maxx = min(l2, m);
  if (minx <= maxx) return maxx - minx + 1;
  return 0;
}
long long f(long long leftMin, long long leftMax, long long rightMin,
            long long rightMax, long long n1, long long n2, long long m1,
            long long m2) {
  return g(leftMin, leftMax, n1, n2) * g(rightMin, rightMax, m1, m2);
}
int main() {
  cin >> n1 >> n2 >> m1 >> m2 >> k;
  MAXV = max(n2, m2);
  generate(0);
  n = a.size();
  sort(a.begin(), a.end());
  long long sum = 0;
  long long cases = (n2 - n1 + 1) * (m2 - m1 + 1);
  for (auto i = 0; i < n - k + 1; i++) {
    long long leftMax = a[i], rightMin = a[i + k - 1];
    long long leftMin = i == 0 ? 1 : a[i - 1] + 1;
    leftMin = max(leftMin, min(n1, m1));
    long long rightMax = (i + k) == n ? MAXV : a[i + k] - 1;
    rightMax = min(rightMax, max(n2, m2));
    if (rightMin > rightMax || leftMin > leftMax || leftMax > rightMin)
      continue;
    sum += f(leftMin, leftMax, rightMin, rightMax, n1, n2, m1, m2);
    sum += f(leftMin, leftMax, rightMin, rightMax, m1, m2, n1, n2);
    if (k == 1) sum -= f(a[i], a[i], a[i], a[i], n1, n2, m1, m2);
  }
  cout << setprecision(9);
  cout << fixed << ((double)sum / (double)cases);
}