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;
long long kgm, pR, pL, vR, vL;
vector<long long> a;
int k;
void init() {
  kgm = (pR - pL + 1) * (vR - vL + 1);
  for (int i = (1); i <= (9); i++) {
    for (int mask = (0); mask <= ((1 << i) - 1); mask++) {
      long long so = 0;
      for (int j = (0); j <= (i - 1); j++)
        if (mask & (1 << j))
          so = so * 10 + 7;
        else
          so = so * 10 + 4;
      a.push_back(so);
    }
  }
  sort((a).begin(), (a).end());
}
pair<long long, long long> get(long long l, long long r, long long L,
                               long long R) {
  if (r < L || R < l) return pair<long long, long long>(-1, -1);
  return pair<long long, long long>(max(l, L), min(r, R));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> pL >> pR >> vL >> vR >> k;
  init();
  long long res = 0;
  for (int i = (0); i <= (int((a).size()) - 1); i++) {
    if (i + k - 1 >= int((a).size())) break;
    long long l = a[i];
    long long r = a[i + k - 1];
    long long L, R;
    if (i == 0)
      L = 1;
    else
      L = a[i - 1] + 1;
    if (i + k - 1 == int((a).size()) - 1)
      R = 1000000000;
    else
      R = a[i + k - 1 + 1] - 1;
    pair<long long, long long> lef = get(pL, pR, L, l);
    pair<long long, long long> rig = get(vL, vR, r, R);
    if (lef.first != -1 && rig.first != -1)
      res += (lef.second - lef.first + 1) * (rig.second - rig.first + 1);
    lef = get(vL, vR, L, l);
    rig = get(pL, pR, r, R);
    if (lef.first != -1 && rig.first != -1)
      res += (lef.second - lef.first + 1) * (rig.second - rig.first + 1) -
             (k == 1);
  }
  cout << fixed << setprecision(13) << res * 1.0 / kgm;
  return 0;
}