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>
string tos(T a) {
  stringstream ss;
  string ret;
  ss << a;
  ss >> ret;
  return ret;
}
vector<long long> luck;
int llen;
void gen(long long v) {
  if (v > 1000000000) return;
  luck.push_back(v);
  gen(v * 10 + 4);
  gen(v * 10 + 7);
}
long long get(long long a, long long b, long long c, long long d, long long k) {
  long long ret = 0;
  long long na, nb, nc, nd;
  for (int(i) = (0); (i) < (llen); (i)++) {
    int j = i + k - 1;
    if (j >= llen) break;
    na = a;
    nb = b;
    nc = c;
    nd = d;
    if (i > 0) na = max(na, luck[i - 1] + 1);
    nb = min(nb, luck[i]);
    nc = max(nc, luck[j]);
    if (j != llen - 1) nd = min(nd, luck[j + 1] - 1);
    if (na <= nb && nc <= nd) ret += (nb - na + 1) * (nd - nc + 1);
  }
  return ret;
}
int main() {
  long long a, b, c, d, k;
  gen(4);
  gen(7);
  sort(luck.begin(), luck.end());
  llen = luck.size();
  while (cin >> a >> b >> c >> d >> k) {
    long long ret = 0;
    ret = get(a, b, c, d, k) + get(c, d, a, b, k);
    if (k == 1)
      for (int(i) = (0); (i) < (llen); (i)++)
        if (luck[i] >= a && luck[i] <= b && luck[i] >= c && luck[i] <= d) ret--;
    long long tot = (b - a + 1) * (d - c + 1);
    double res = ret / (double)tot;
    printf("%.12lf\n", res);
  }
  return 0;
}