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 x[2111], tp = 0;
void dfs(int pre, int dig, int sum) {
  if (dig == 9) {
    if (!sum) return;
    x[tp++] = sum;
    return;
  }
  if (pre == 0) {
    dfs(0, dig + 1, sum);
  }
  dfs(4, dig + 1, sum * 10 + 4);
  dfs(7, dig + 1, sum * 10 + 7);
}
double turn(int x, int y, int a, int b) {
  int l = max(x, a), r = min(y, b);
  if (l > r)
    return 0.0;
  else
    return r - l + 1;
}
int main() {
  int l, r, a, b, k;
  scanf("%d %d %d %d %d", &l, &r, &a, &b, &k);
  x[tp++] = 0;
  x[tp++] = 1100000000;
  dfs(0, 0, 0);
  sort(x, x + tp);
  double ans = 0;
  for (int i = 1; i + k - 1 < tp - 1; i++) {
    if (k != 1)
      ans += turn(x[i - 1] + 1, x[i], l, r) *
                 turn(x[i + k - 1], x[i + k] - 1, a, b) +
             turn(x[i - 1] + 1, x[i], a, b) *
                 turn(x[i + k - 1], x[i + k] - 1, l, r);
    else {
      ans += turn(x[i - 1] + 1, x[i], l, r) *
                 turn(x[i + k - 1], x[i + k] - 1, a, b) +
             turn(x[i - 1] + 1, x[i], a, b) *
                 turn(x[i + k - 1], x[i + k] - 1, l, r);
      if (x[i] >= l && x[i] <= r && x[i] >= a && x[i] <= b) ans -= 1;
    }
  }
  printf("%.10f\n", ans / (1.0 * (r - l + 1) * (b - a + 1)));
  return 0;
}