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 num[10001], nc = 0;
inline void dfs(int a) {
  num[nc++] = a;
  if (a > 100000000) return;
  dfs(10 * a + 4);
  dfs(10 * a + 7);
}
inline long long f(int l, int r, int x, int y) {
  l = max(l, x);
  r = min(r, y);
  if (r < l) return 0;
  return r - l + 1ll;
}
int main() {
  dfs(0);
  sort(num, num + nc);
  num[nc++] = 1000000002;
  int pl, pr, vl, vr, k;
  scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &k);
  long long ans = 0;
  for (int i = 1; i + k - 1 < nc - 1; i++) {
    int lmi, lma, rmi, rma;
    lmi = num[i - 1] + 1;
    lma = num[i];
    rmi = num[i + k - 1];
    rma = num[i + k] - 1;
    long long k1 = f(lmi, lma, pl, pr) * f(rmi, rma, vl, vr);
    long long k2 = f(lmi, lma, vl, vr) * f(rmi, rma, pl, pr);
    ans += k1 + k2;
    if (k1 && k2 && k == 1) ans--;
  }
  printf("%.12lf\n", (double)ans / (vr - vl + 1ll) / (pr - pl + 1ll));
}