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;
int maxn = -(int)1e9;
set<long long> lucky;
void go(long long x) {
  if (x > maxn) return;
  if (x) lucky.insert(x);
  go(10 * x + 4), go(10 * x + 7);
}
long long query(int l, int r, pair<int, int> p, pair<int, int> v, int pl,
                int nv) {
  pl = max(pl + 1, p.first);
  nv = min(nv - 1, v.second);
  l = min(l, p.second);
  r = max(r, v.first);
  if (l < pl || nv < r) return 0;
  return 1LL * (l - pl + 1) * (nv - r + 1);
}
int main() {
  pair<int, int> p, v;
  int k;
  cin >> p.first >> p.second >> v.first >> v.second >> k;
  maxn = max(maxn, max(p.second, v.second));
  go(0);
  vector<long long> d((lucky).begin(), (lucky).end());
  if (p > v) swap(p, v);
  int pl = -1011111111;
  long long total = 0;
  for (int i = 0; i + k - 1 < d.size(); i++) {
    int l = (int)d[i], r = (int)d[i + k - 1];
    int nv = i + k < d.size() ? d[i + k] : 1011111111;
    total += query(l, r, p, v, pl, nv);
    total += query(l, r, v, p, pl, nv);
    pl = l;
  }
  if (k == 1) {
    for (int _n(d.size()), i(0); i < _n; i++) {
      if (d[i] < p.first || d[i] > p.second || d[i] < v.first ||
          d[i] > v.second)
        continue;
      total--;
    }
  }
  double ans =
      1.0 * total / (p.second - p.first + 1) / (v.second - v.first + 1);
  printf("%.10f\n", ans);
  return 0;
}