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;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  int fh = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= fh;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
int pl, pr, vl, vr, k;
vector<int> lucky;
long double ans = 0;
void dfs(long long x) {
  if (x > 1000000000) return;
  if (x > 0) lucky.push_back(x);
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
signed main() {
  cin >> pl >> pr >> vl >> vr >> k;
  dfs(0LL);
  sort(lucky.begin(), lucky.end());
  for (int lid = 0; lid < lucky.size(); ++lid) {
    int rid = lid + k - 1;
    if (rid >= lucky.size()) continue;
    int lp = max(pl, (lid == 0 ? 1 : lucky[lid - 1] + 1)),
        rp = min(pr, lucky[lid]);
    int lv = max(vl, lucky[rid]),
        rv = min(vr,
                 (rid == lucky.size() - 1 ? 1000000007 : lucky[rid + 1] - 1));
    ans += (long double)max(0, rp - lp + 1) / (long double)(pr - pl + 1) *
           (long double)max(0, rv - lv + 1) / (long double)(vr - vl + 1);
    lv = max(vl, (lid == 0 ? 1 : lucky[lid - 1] + 1)), rv = min(vr, lucky[lid]);
    lp = max(pl, lucky[rid]),
    rp = min(pr, (rid == lucky.size() - 1 ? 1000000007 : lucky[rid + 1] - 1));
    ans += (long double)max(0, rp - lp + 1) / (long double)(pr - pl + 1) *
           (long double)max(0, rv - lv + 1) / (long double)(vr - vl + 1);
  }
  for (int i = 0; i < lucky.size(); ++i)
    if (k == 1 && pl <= lucky[i] && lucky[i] <= pr && vl <= lucky[i] &&
        lucky[i] <= vr)
      ans -= (long double)1 / (long double)(pr - pl + 1) /
             (long double)(vr - vl + 1);
  cout << fixed << setprecision(12) << ans << endl;
  return 0;
}