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 tab[5005];
queue<long long> Q;
int main() {
  long long pl, pr, vl, vr, K;
  scanf("%I64d%I64d%I64d%I64d%I64d", &vl, &vr, &pl, &pr, &K);
  long long N = 1e11, id = 3;
  tab[1] = 4;
  tab[2] = 7;
  Q.push(4);
  Q.push(7);
  while (!Q.empty()) {
    long long u = Q.front();
    Q.pop();
    if (u * 10 + 4 > N) continue;
    tab[id++] = u * 10 + 4;
    tab[id++] = u * 10 + 7;
    Q.push(u * 10 + 4);
    Q.push(u * 10 + 7);
  }
  long long ans = 0;
  for (int i = 1;; ++i) {
    if (tab[i - 1] + 1 > vr && tab[i - 1] + 1 > pr) break;
    int nx = i + K - 1;
    long long l1 = max(pl, tab[i - 1] + 1);
    long long r1 = min(pr, tab[i]);
    long long l2 = max(vl, tab[nx]);
    long long r2 = min(vr, tab[nx + 1] - 1);
    if (l1 <= r1 && l2 <= r2) ans += 1LL * (r1 - l1 + 1) * (r2 - l2 + 1);
    l1 = max(vl, tab[i - 1] + 1);
    r1 = min(vr, tab[i]);
    l2 = max(pl, tab[nx]);
    r2 = min(pr, tab[nx + 1] - 1);
    if (l1 <= r1 && l2 <= r2) ans += (r1 - l1 + 1) * (r2 - l2 + 1);
  }
  if (K == 1) {
    int l = max(vl, pl), r = min(vr, pr);
    for (int i = 1; i < id; ++i)
      if (l <= tab[i] && tab[i] <= r) --ans;
  }
  printf("%.15f\n", 1.0 * ans / (pr - pl + 1) / (vr - vl + 1));
  return 0;
}