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 a[10000];
long long sum = 0;
void dfs(long long x) {
  if (x > 5e9) return;
  a[sum++] = x;
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
long long check(long long l1, long long r1, long long l2, long long r2) {
  long long l = max(l1, l2);
  long long r = min(r1, r2);
  long long ans = (r - l + 1);
  if (ans < 0) ans = 0;
  return ans;
}
int main() {
  long long x1, x2, y2, y1, k;
  cin >> x1 >> y1 >> x2 >> y2 >> k;
  dfs(0);
  sort(a, a + sum);
  long long l = 1, r = k;
  double ans = 0;
  long long sum1 = y1 - x1 + 1;
  long long sum2 = y2 - x2 + 1;
  while (r <= sum) {
    long long cnt1_4 = check(a[l - 1] + 1, a[l], x1, y1);
    long long cnt1_7 = check(a[r], a[r + 1] - 1, x1, y1);
    long long cnt2_4 = check(a[l - 1] + 1, a[l], x2, y2);
    long long cnt2_7 = check(a[r], a[r + 1] - 1, x2, y2);
    ans += cnt1_4 * 1.0 / sum1 * cnt2_7 * 1.0 / sum2;
    ans += cnt2_4 * 1.0 / sum2 * cnt1_7 * 1.0 / sum1;
    l++, r++;
  }
  if (k == 1)
    for (int i = 1; i <= sum; i++)
      if (check(a[i], a[i], x1, y1) && check(a[i], a[i], x2, y2))
        ans -= 1.0 / sum1 / sum2;
  printf("%.10f\n", fabs(ans));
}