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> v;
void dfs(long long x, int num) {
  if (x != 0) v.push_back(x);
  if (num == 9) return;
  dfs(x * 10 + 4, num + 1);
  dfs(x * 10 + 7, num + 1);
}
pair<long long, long long> L[1024], R[1024];
pair<long long, long long> seg(pair<long long, long long> p,
                               pair<long long, long long> q) {
  if (p.first > p.second || q.first > q.second) return make_pair(-11LL, -11LL);
  long long mini = max(q.first, p.first);
  long long maxi = min(q.second, p.second);
  if (mini > maxi)
    return make_pair(-11LL, -11LL);
  else
    return make_pair(mini, maxi);
}
int main() {
  long long pl, pr, vr, vl;
  int k;
  cin >> pl >> pr >> vl >> vr >> k;
  dfs(0, 0);
  sort(v.begin(), v.end());
  long long ans = 0;
  for (int i = 0; i < (int)v.size(); i++) {
    if (i != 0)
      L[i] = make_pair(v[i] - (v[i] - v[i - 1]) + 1, v[i]);
    else
      L[i] = make_pair(-11LL, v[i]);
    if (i + 1 != (int)v.size())
      R[i] = make_pair(v[i], v[i] + (v[i + 1] - v[i]) - 1);
    else
      R[i] = make_pair(v[i], 200000000000000LL);
  }
  for (int i = 0; i < (int)v.size(); i++) {
    for (int j = 0; j < (int)v.size(); j++) {
      pair<long long, long long> p = seg(L[i], make_pair(pl, pr));
      pair<long long, long long> q = seg(R[j], make_pair(vl, vr));
      if ((make_pair(-11LL, -11LL) != p) && (make_pair(-11LL, -11LL) != q) &&
          (i + k - 1 == j)) {
        ans += (p.second - p.first + 1) * (q.second - q.first + 1);
      }
      p = seg(R[i], make_pair(pl, pr));
      q = seg(L[j], make_pair(vl, vr));
      if ((make_pair(-11LL, -11LL) != p) && (make_pair(-11LL, -11LL) != q) &&
          (i - k + 1 == j)) {
        ans += (p.second - p.first + 1) * (q.second - q.first + 1);
      }
    }
  }
  if (k == 1) {
    for (int i = 0; i < (int)v.size(); i++) {
      if (pl <= v[i] && v[i] <= pr && vl <= v[i] && v[i] <= vr) ans--;
    }
  }
  long long tot = (vr - vl + 1) * (pr - pl + 1);
  printf("%.10f\n", (double)ans / (double)tot);
}