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;
using ll = long long;
ll cnt1e[3333], cnt2e[3333], cnt1i[3333], cnt2i[3333];
set<int> lk;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int len = 1; len <= 9; len++) {
    for (int x = 0; x < (1 << len); x++) {
      int res = 0;
      for (int i = 0; i < len; i++) {
        if (x & (1 << i))
          res = res * 10 + 4;
        else
          res = res * 10 + 7;
      }
      lk.insert(res);
    }
  }
  lk.insert(1000000001);
  int amt = lk.size();
  int pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  int pre = 0, cur = 0;
  for (auto R : lk) {
    int L = pre;
    cnt1i[cur] += max(0, min(pr, R - 1) - max(pl, L) + 1);
    cnt1e[cur] += max(0, min(pr, R) - max(pl, L + 1) + 1);
    pre = R;
    cur++;
  }
  pre = 0, cur = 0;
  for (auto R : lk) {
    int L = pre;
    cnt2i[cur] += max(0, min(vr, R - 1) - max(vl, L) + 1);
    cnt2e[cur] += max(0, min(vr, R) - max(vl, L + 1) + 1);
    pre = R;
    cur++;
  }
  ll tot = 0;
  for (int i = 0; i <= amt; i++) {
    tot += cnt1e[i] * cnt2i[i + k];
    tot += cnt2e[i] * cnt1i[i + k];
  }
  if (k == 1) {
    for (auto x : lk)
      if (x >= vl && x <= vr && x >= pl && x <= pr) tot--;
  }
  cout << fixed << setprecision(12)
       << (double)tot / (double)((ll)(vr - vl + 1) * (ll)(pr - pl + 1)) << '\n';
}