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 N = 20, Mod = 1e9 + 7;
int pl, pr, vl, vr, k;
long double p, q;
vector<int> v;
void count(int t) {
  int L, R;
  L = min(v[t], pr) - max(v[t - 1] + 1, pl);
  R = min(v[t + k] - 1, vr) - max(v[t + k - 1], vl);
  p += max(L + 1, 0) * 1ll * max(R + 1, 0);
  L = min(v[t], vr) - max(v[t - 1] + 1, vl);
  R = min(v[t + k] - 1, pr) - max(v[t + k - 1], pl);
  p += max(L + 1, 0) * 1ll * max(R + 1, 0);
  if (k == 1) {
    if (vl <= v[t] && v[t] <= vr && pl <= v[t] && v[t] <= pr) p--;
  }
}
void add(int length) {
  for (int i = 0; i < (1 << length); i++) {
    int num = 0;
    for (int j = 0; j < length; j++) {
      num *= 10;
      if (i & (1 << j))
        num += 4;
      else
        num += 7;
    }
    v.push_back(num);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> pl >> pr >> vl >> vr >> k;
  q = (pr - pl + 1) * 1ll * (vr - vl + 1);
  for (int i = 1; i <= 9; i++) add(i);
  v.push_back(0);
  v.push_back(Mod);
  v.push_back(Mod);
  sort(v.begin(), v.end());
  int n = v.size();
  for (int i = 1; i < n - k - 1; i++) {
    count(i);
  }
  cout << fixed << setprecision(10) << p / q;
  return 0;
}