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 int inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
pair<int, int> p, v;
int k;
vector<int> happy;
void go(int x) {
  happy.push_back(x);
  if (10ll * x + 4 <= inf) go(10 * x + 4);
  if (10ll * x + 7 <= inf) go(10 * x + 7);
}
double calc(pair<int, int> p, int l, int r) {
  double sz = p.second - p.first + 1;
  int L = max(l, p.first), R = min(r, p.second);
  int ans = max(0, R - L + 1);
  return ans / sz;
}
double calc(int l1, int r1, int l2, int r2) {
  if (l1 > r1) return 0.0;
  if (l2 > r2) return 0.0;
  return calc(p, l1, r1) * calc(v, l2, r2) + calc(v, l1, r1) * calc(p, l2, r2);
}
int main() {
  go(0);
  cin >> p.first >> p.second >> v.first >> v.second >> k;
  sort(happy.begin(), happy.end());
  double ans = 0;
  for (int i = 1; i + k <= happy.size(); i++) {
    ans += calc((i ? happy[i - 1] + 1 : 1), happy[i], happy[i + k - 1],
                (i + k < happy.size() ? happy[i + k] - 1 : inf));
  }
  if (k == 1) {
    for (int i = 1; i + 1 <= happy.size(); i++) {
      ans -= calc(happy[i], happy[i], happy[i], happy[i]) / 2.0;
    }
  }
  printf("%.9f\n", ans);
  return 0;
}