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<int> v;
inline void dfs(int cur) {
  v.push_back(cur);
  if (10ll * cur + 4 >= 1e9) return;
  dfs(10 * cur + 4);
  dfs(10 * cur + 7);
}
int l1, r1, l2, r2, k;
inline int getlen(int l1, int r1, int l2, int r2) {
  int l = max(l1, l2), r = min(r1, r2);
  return max(r - l + 1, 0);
}
int main() {
  dfs(0);
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  k--;
  sort(v.begin(), v.end());
  int s = v.size();
  v.push_back(1e9);
  long long ans = 0;
  for (int i = 1; i + k < s; i++)
    ans += 1ll * getlen(l1, r1, v[i - 1] + 1, v[i]) *
           getlen(l2, r2, v[i + k], v[i + k + 1] - 1);
  for (int i = 1; i + k < s; i++)
    ans += 1ll * getlen(l2, r2, v[i - 1] + 1, v[i]) *
           getlen(l1, r1, v[i + k], v[i + k + 1] - 1);
  if (k == 0)
    for (int i = 1; i < s; i++)
      if (getlen(l1, r1, v[i], v[i]) && getlen(l2, r2, v[i], v[i])) ans--;
  printf("%.15f", 1.0 * ans / (r1 - l1 + 1) / (r2 - l2 + 1));
}