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 oo = (int)1e9;
const long double PI = 2 * acos(0.0);
const long double eps = 1e-9;
vector<long long> lucky;
void gen(int d1, int d2) {
  for (int i = 1; i < (int)((1 << 9)); i++) {
    long long pw = 1, num = 0, t = i;
    while (t) {
      if (t % 2 == 0)
        num += d1 * pw;
      else
        num += d2 * pw;
      pw *= 10;
      t /= 2;
    }
    lucky.push_back(num);
  }
}
long long s1, e1, s2, e2, K;
long long calc(int a, int b) {
  long long ret = 0;
  long long prev = 1, after = oo;
  if (a != 0) prev = lucky[a - 1] + 1;
  if (b != ((long long)lucky.size()) - 1) after = lucky[b + 1] - 1;
  bool f1 = false, f2 = false;
  if (lucky[a] >= s1 && prev <= e1 && after >= s2 && lucky[b] <= e2) {
    ret += (min(lucky[a], e1) - max(s1, prev) + 1) *
           (min(after, e2) - max(lucky[b], s2) + 1);
    f1 = true;
  }
  if (lucky[a] >= s2 && prev <= e2 && after >= s1 && lucky[b] <= e1) {
    ret += (min(lucky[a], e2) - max(s2, prev) + 1) *
           (min(after, e1) - max(lucky[b], s1) + 1);
    f2 = true;
  }
  if (K == 1 && f1 && f2) ret--;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  gen(4, 7);
  gen(7, 4);
  sort(lucky.begin(), lucky.end());
  cin >> s1 >> e1 >> s2 >> e2 >> K;
  long long tot = (e1 - s1 + 1) * (e2 - s2 + 1), val = 0;
  for (int i = K - 1; i < (int)(((long long)lucky.size())); i++) {
    val += calc(i - K + 1, i);
  }
  cout << fixed << setprecision(9) << double(val) / tot << endl;
  return 0;
}