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;
long long p1, p2, v1, v2, k, MAX = 1000 * 1000 * 1000;
double ans;
vector<long long> lucky;
void gen(long long cur) {
  if (cur > MAX) return;
  lucky.push_back(cur);
  gen(cur * 10 + 4);
  gen(cur * 10 + 7);
}
double count(long long st, long long en, long long ind) {
  if (ind == 0) {
    if (st > p2 or p1 > en) return 0;
    return (min(p2, en) - max(p1, st) + 1) / double(p2 - p1 + 1);
  } else {
    if (st > v2 or v1 > en) return 0;
    return (min(v2, en) - max(v1, st) + 1) / double(v2 - v1 + 1);
  }
}
void add(long long st, long long en) {
  long long st1, en1;
  if (st == 0)
    st1 = 0;
  else
    st1 = lucky[st - 1] + 1;
  if (en == lucky.size() - 1)
    en1 = MAX;
  else
    en1 = lucky[en + 1] - 1;
  st = lucky[st];
  en = lucky[en];
  ans += count(st1, st, 0) * count(en, en1, 1);
  ans += count(st1, st, 1) * count(en, en1, 0);
  if (st == en and st >= p1 and st <= p2 and st >= v1 and st <= v2)
    ans -= 1.0 / ((p2 - p1 + 1) * (v2 - v1 + 1));
}
int main() {
  cin >> p1 >> p2;
  cin >> v1 >> v2;
  cin >> k;
  gen(4);
  gen(7);
  sort(lucky.begin(), lucky.end());
  for (long long i = 0; i + k - 1 < lucky.size(); i++) add(i, i + k - 1);
  cout.precision(9);
  cout << fixed << ans << endl;
  return 0;
}