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 long long mod = 1e9 + 9;
const int maxn = 3e5 + 50;
const long long inf = 0x3f3f3f3f3f3f3f3fLL;
long long has[maxn];
int num = 0;
long long get(long long al, long long ar, long long ql, long long qr) {
  long long l = max(al, ql);
  long long r = min(ar, qr);
  return l <= r ? (r - l + 1LL) : 0;
}
vector<long long> pre, nex;
void init() {
  pre.push_back(0);
  for (int i = 0; i < 10; i++) {
    nex.clear();
    for (auto i : pre) {
      long long now = i * 10LL + 4LL;
      nex.push_back(now);
      has[++num] = now;
      now = i * 10LL + 7LL;
      nex.push_back(now);
      has[++num] = now;
    }
    pre = nex;
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  init();
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  long long down = 1LL * (pr - pl + 1LL) * (vr - vl + 1LL);
  long long ans = 0;
  for (long long i = 1; i <= 2040; i++) {
    long long pre = i, next = i + k - 1;
    ans += get(has[pre - 1] + 1, has[pre], pl, pr) *
           get(has[next], has[next + 1] - 1, vl, vr);
    ans += get(has[pre - 1] + 1, has[pre], vl, vr) *
           get(has[next], has[next + 1] - 1, pl, pr);
  }
  if (k == 1) {
    for (int i = 1; i <= 2040; i++) {
      ans -= get(has[i], has[i], pl, pr) * get(has[i], has[i], vl, vr);
    }
  }
  cout << fixed;
  cout << setprecision(13) << double(ans * 1.0 / down * 1.0) << endl;
  return 0;
}