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 N = 2e6 + 5, Mod = 1e9 + 7, MAXN = 1e8 + 1000, LG = 30,
                M = 1e6 + 5, P = 727, Sq = 100;
const long long inf = 3e18 + 10;
vector<long long> v;
long double pl, pr, vl, vr, k;
long double check1(long double f1, long double f2) {
  if (f2 < pl || f1 > pr) return 0;
  f1 = max(f1, pl);
  f2 = min(f2, pr);
  if (f1 > f2) return 0;
  long double tt = (f2 - f1 + 1), pp = pr - pl + 1;
  tt /= pp;
  return tt;
}
long double check2(long double f1, long double f2) {
  if (f2 < vl || f1 > vr) return 0;
  f1 = max(f1, vl);
  f2 = min(f2, vr);
  if (f1 > f2) return 0;
  long double tt = (f2 - f1 + 1), pp = vr - vl + 1;
  tt /= pp;
  return tt;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> pl >> pr >> vl >> vr >> k;
  for (long long i = 1; i <= 11; i++) {
    for (long long j = 0; j < (1 << i); j++) {
      long long ret = 0;
      for (long long x = 0; x < i; x++) {
        if (((1 << x) & j) > 0)
          ret *= 10, ret += 4;
        else
          ret *= 10, ret += 7;
      }
      v.push_back(ret);
    }
  }
  sort(v.begin(), v.end());
  long double res = 0;
  for (long long i = k - 1; i < (long long)v.size() - 1; i++) {
    long long l2 = v[i - k + 1], l1 = 0;
    if (i == k - 1)
      l1 = 1;
    else
      l1 = v[i - k] + 1;
    long long r2 = v[i + 1] - 1, r1 = v[i];
    res += check1(l1, l2) * check2(r1, r2);
    res += check2(l1, l2) * check1(r1, r2);
    if (l1 > r1) swap(l1, r1), swap(l2, r2);
    long long bad1 = max(l1, r1), bad2 = min(l2, r2);
    if (bad2 < bad1) continue;
    long double fg = vr - vl + 1;
    res -= check1(bad1, bad2) / fg;
  }
  cout << fixed << setprecision(12) << res << endl;
  return 0;
}