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;
vector<long long> vec;
void solve(long long x) {
  long long o = x * 10 + 4;
  long long t = x * 10 + 7;
  if (o <= 1e9) {
    vec.push_back(o);
    solve(o);
  }
  if (t <= 1e9) {
    vec.push_back(t);
    solve(t);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve(0);
  vec.push_back(0);
  vec.push_back(1e9 + 1);
  sort(vec.begin(), vec.end());
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  long long tot = (pr - pl + 1) * (vr - vl + 1);
  long long curr = 0;
  for (int i = 1; i < vec.size() - 1; i++) {
    int nxt = i + k - 1;
    if (nxt >= vec.size() - 1) break;
    long long L1 = vec[i - 1] + 1, R1 = vec[i];
    long long L2 = vec[nxt], R2 = vec[nxt + 1] - 1;
    L1 = max(L1, pl);
    R1 = min(R1, pr);
    L2 = max(L2, vl);
    R2 = min(R2, vr);
    long long cnt1 = (R1 - L1 + 1), cnt2 = (R2 - L2 + 1);
    long long LL1 = 0, RR1 = 0;
    long long LL2 = 0, RR2 = 0;
    if (cnt1 > 0 && cnt2 > 0) {
      curr += cnt1 * cnt2;
      LL1 = L1;
      RR1 = R1;
      LL2 = L2;
      RR2 = R2;
    }
    L1 = vec[i - 1] + 1;
    R1 = vec[i];
    L2 = vec[nxt];
    R2 = vec[nxt + 1] - 1;
    L1 = max(L1, vl);
    R1 = min(R1, vr);
    L2 = max(L2, pl);
    R2 = min(R2, pr);
    cnt1 = (R1 - L1 + 1);
    cnt2 = (R2 - L2 + 1);
    if (cnt1 > 0 && cnt2 > 0) {
      curr += cnt1 * cnt2;
      if (RR1 != 0) {
        LL1 = max(LL1, L2);
        RR1 = min(RR1, R2);
        LL2 = max(LL2, L1);
        RR2 = min(RR2, R1);
        long long cc1 = (RR1 - LL1 + 1);
        long long cc2 = (RR2 - LL2 + 1);
        if (cc1 > 0 && cc2 > 0) curr -= cc1 * cc2;
      }
    }
  }
  cout << fixed << setprecision(9) << (long double)(curr) / (long double)(tot);
  return 0;
}