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) {
  if (x > 1000000000) return;
  v.push_back(x);
  dfs(10 * x + 4);
  dfs(10 * x + 7);
}
long long func(long long x1, long long x2, long long y1, long long y2, int K) {
  int i, j;
  int sz = v.size();
  long long ans = 0;
  for ((i) = 0; (i) < (int)(sz); (i)++) {
    j = i + K - 1;
    if (j >= sz) break;
    long long xlow = x1, xhigh = x2, ylow = y1, yhigh = y2;
    if (i != 0) xlow = max(xlow, v[i - 1] + 1);
    xhigh = min(xhigh, v[i]);
    ylow = max(ylow, v[j]);
    if (j != sz - 1) yhigh = min(yhigh, v[j + 1] - 1);
    if (xhigh >= xlow && yhigh >= ylow)
      ans += (xhigh - xlow + 1) * (yhigh - ylow + 1);
  }
  return ans;
}
int main(void) {
  int i;
  dfs(4);
  dfs(7);
  sort(v.begin(), v.end());
  int sz = v.size();
  long long x1, y1, x2, y2;
  int K;
  cin >> x1 >> x2 >> y1 >> y2 >> K;
  long long total = (x2 - x1 + 1) * (y2 - y1 + 1);
  long long tmp = func(x1, x2, y1, y2, K) + func(y1, y2, x1, x2, K);
  if (K == 1)
    for ((i) = 0; (i) < (int)(sz); (i)++)
      if (v[i] >= x1 && v[i] <= x2 && v[i] >= y1 && v[i] <= y2) tmp--;
  double ans = (double)tmp / total;
  printf("%.12f\n", ans);
  return 0;
}