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;
int pL, pR, vL, vR, k;
vector<int> v;
void f(long long x) {
  if (x > 1e9) return;
  v.push_back(x);
  f(x * 10 + 4);
  f(x * 10 + 7);
}
int get(int L, int R, int st, int en) {
  if (L > R || (L > en || R < st)) return 0;
  if (st >= L && en <= R) return en - st + 1;
  int ret = min(R, en) - max(L, st) + 1;
  assert(ret >= 0);
  return ret;
}
int main() {
  f(0);
  int n = v.size();
  v.push_back(1e9 + 1);
  sort(((v).begin()), ((v).end()));
  cin >> pL >> pR >> vL >> vR >> k;
  long double ret = 0;
  for (int i = 1; i + k - 1 < n; i++) {
    long long x = get(v[i - 1] + 1, v[i], pL, pR) * 1. *
                  get(v[i + k - 1], v[i + k] - 1, vL, vR);
    long long y = get(v[i - 1] + 1, v[i], vL, vR) * 1. *
                  get(v[i + k - 1], v[i + k] - 1, pL, pR);
    if (k == 1 && get(pL, pR, vL, vR) && v[i] >= max(pL, vL) &&
        v[i] <= min(pR, vR))
      x--;
    ret += x + y;
  }
  cout << setprecision(12) << fixed;
  cout << ret * 1. / (pR - pL + 1) / (vR - vL + 1) << '\n';
  return 0;
}