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 GetArray(vector<T> &vec, int n) {
  for (int i = 0; i < n; ++i) {
    T x;
    cin >> x;
    vec.push_back(x);
  }
}
const int inf = 2 * 1e9 + 5;
int pl, pr, vl, vr, k;
vector<long long> kvec;
void Generate(long long val, int bit) {
  if (bit > 9) return;
  if (bit > 0) kvec.push_back(val);
  Generate(val * 10 + 4, bit + 1);
  Generate(val * 10 + 7, bit + 1);
}
long long intersect(long long l1, long long r1, long long l2, long long r2) {
  if (l1 > r2 || l2 > r1) return 0;
  return min(r1, r2) - max(l1, l2) + 1;
}
int Main() {
  cin >> pl >> pr >> vl >> vr >> k;
  Generate(0, 0);
  sort(kvec.begin(), kvec.end());
  long long tot = 0;
  for (int k1 = 0, k2 = k1 + (k - 1); k2 < kvec.size(); ++k1, ++k2) {
    long long l = k1 - 1 >= 0 ? kvec[k1 - 1] + 1 : 1;
    long long r = k2 + 1 < kvec.size() ? kvec[k2 + 1] - 1 : 1e9;
    tot += intersect(l, kvec[k1], pl, pr) * intersect(kvec[k2], r, vl, vr);
    tot += intersect(l, kvec[k1], vl, vr) * intersect(kvec[k2], r, pl, pr);
    if (k == 1) {
      if (pl <= kvec[k1] && kvec[k1] <= pr && vl <= kvec[k1] && kvec[k1] <= vr)
        --tot;
    }
  }
  double ans = 1.0 * tot / (1LL * (pr - pl + 1) * (vr - vl + 1));
  cout << ans << endl;
  return 0;
}
int main() {
  cout.precision(20);
  Main();
  return 0;
}