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;
inline void err(istringstream *iss) {}
template <typename T, typename... Args>
void err(istringstream *iss, const T &$val, const Args &...args) {
  string $name;
  *iss >> $name;
  if ($name.back() == ',') $name.pop_back();
  cout << $name << " = " << $val << "; ", err(iss, args...);
}
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const ll llinf = (1ll << 61) - 1;
int n, m, K, T, Q, cn;
int lucky[1100], li, pl, pr, vl, vr;
ll common(const pair<int, int> &o, const pair<int, int> &p) {
  return max(0, min(o.second, p.second) - max(o.first, p.first) + 1);
}
int main() {
  cin.tie(NULL);
  lucky[li++] = 0;
  for (int i = 1; i <= 9; i++) {
    for (int j = 0; j < 1 << i; j++) {
      int num = 0;
      for (int k = i - 1; k >= 0; k--) {
        num = num * 10 + ((j >> k) & 1 ? 7 : 4);
      }
      lucky[li++] = num;
    }
  }
  lucky[li++] = 1000000007;
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &K);
  if (!(pl < vl or (pl == vl and pr < vr))) swap(pl, vl), swap(pr, vr);
  ll cnt = 0;
  for (int i = 1; i + K - 1 < li - 1; i++) {
    int l = i, r = i + K - 1;
    cnt += common({lucky[l - 1] + 1, lucky[l]}, {pl, pr}) *
           common({lucky[r], lucky[r + 1] - 1}, {vl, vr});
    cnt += common({lucky[l - 1] + 1, lucky[l]}, {vl, vr}) *
           common({lucky[r], lucky[r + 1] - 1}, {pl, pr});
    if (K == 1)
      cnt -= common({lucky[l], lucky[l]}, {vl, vr}) *
             common({lucky[r], lucky[r]}, {pl, pr});
  }
  printf("%.12f\n", cnt * 1.0 / ((pr - pl + 1) * ll(vr - vl + 1)));
  return 0;
}