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;
const double inf = 1 << 30;
int k;
double p1, p2, l1, l2;
vector<double> q;
void dfs(int a, int b, double num) {
  if (a == 0 && b == 0) {
    q.push_back(num);
    return;
  }
  if (a > 0) dfs(a - 1, b, num * 10.0 + 4.0);
  if (b > 0) dfs(a, b - 1, num * 10.0 + 7.0);
}
int main() {
  q.clear();
  int i, j;
  for (i = 1; i < 10; i++)
    for (j = 0; j <= i; j++) dfs(j, i - j, 0);
  q.push_back(0.0);
  q.push_back(inf);
  sort(q.begin(), q.end());
  while (scanf("%lf%lf%lf%lf%d", &p1, &p2, &l1, &l2, &k) == 5) {
    int s1, s2, r1, r2;
    for (i = 0; i < q.size(); i++) {
      if (q[i] < p1) s1 = i;
      if (q[i] > p2) {
        s2 = i;
        break;
      }
    }
    for (i = 0; i < q.size(); i++) {
      if (q[i] < l1) r1 = i;
      if (q[i] > l2) {
        r2 = i;
        break;
      }
    }
    double sum = 0;
    double ret = (p2 - p1 + 1) * (l2 - l1 + 1);
    for (i = s1; i < s2; i++)
      for (j = r1; j < r2; j++) {
        if (j - i == k) {
          double ma, mi;
          ma = min(q[i + 1], p2);
          mi = max(q[i], p1);
          double tmp = ma - mi;
          if (i == s1) tmp++;
          ma = min(q[j + 1], l2);
          mi = max(q[j], l1);
          double res = ma - mi;
          if (j == r2 - 1) res++;
          sum += tmp * res;
        } else if (i - j == k) {
          double ma, mi;
          ma = min(q[i + 1], p2);
          mi = max(q[i], p1);
          double tmp = ma - mi;
          if (i == s2 - 1) tmp++;
          ma = min(q[j + 1], l2);
          mi = max(q[j], l1);
          double res = ma - mi;
          if (j == r1) res++;
          sum += tmp * res;
        }
      }
    if (k == 1)
      for (i = s1 + 1; i < s2; i++)
        for (j = r1 + 1; j < r2; j++)
          if (i == j) sum--;
    printf("%.12f\n", sum / ret);
  }
  return 0;
}