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;
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
const double pi = asin(1.0) * 2;
int n;
int m;
long long pl, pr, vl, vr;
int k;
vector<long long> h;
double prob(long long l, long long r, long long L, long long R) {
  L = max(l, L);
  R = min(R, r);
  if (L > R) return 0.0;
  return (R - L + 1) / (double)(r - l + 1);
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  h.push_back(0);
  int cur = 0;
  for (int i = cur; h[i] * 10 + 4 <= 1e10; i++) {
    h.push_back(h[i] * 10 + 4);
    h.push_back(h[i] * 10 + 7);
  }
  sort((h).begin(), (h).end());
  double ans = 0;
  for (int i = 1; i < (int)((h).size()); i++) {
    if (i + k >= (int)((h).size())) break;
    long long l1 = h[i - 1] + 1, r1 = h[i];
    long long l2 = h[i + k - 1], r2 = h[i + k] - 1;
    ans += prob(pl, pr, l1, r1) * prob(vl, vr, l2, r2);
    ans += prob(pl, pr, l2, r2) * prob(vl, vr, l1, r1);
    ans -= prob(pl, pr, max(l2, l1), min(r1, r2)) *
           prob(vl, vr, max(l2, l1), min(r1, r2));
  }
  printf("%.10lf\n", ans);
  return 0;
}