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<long long> lucky;
void cal(long long value) {
  if (value < 1000000000) lucky.push_back(value);
  if (value >= 1000000000) return;
  value *= 10;
  cal(value + 4);
  cal(value + 7);
}
int al, ar, bl, br, k;
int get_intersect(int u, int v, int l, int r) {
  u = max(u, l);
  v = min(v, r);
  return (v >= u) ? v - u + 1 : 0;
}
double solve() {
  long long Count = 0, a, b, c, d;
  for (int i = 0; i < lucky.size(); i++) {
    int j = i + k - 1;
    if (j >= lucky.size()) break;
    a = get_intersect(al, ar, lucky[i - 1] + 1, lucky[i]);
    b = get_intersect(bl, br, lucky[j], lucky[j + 1] - 1);
    Count += a * b;
    c = get_intersect(bl, br, lucky[i - 1] + 1, lucky[i]);
    d = get_intersect(al, ar, lucky[j], lucky[j + 1] - 1);
    Count += c * d;
    if (k == 1 && a && b && c && d) Count--;
  }
  a = ar - al + 1;
  b = br - bl + 1;
  return (double)Count / a / b;
}
int main() {
  lucky.push_back(-1);
  cal(4);
  cal(7);
  sort(lucky.begin(), lucky.end());
  lucky.push_back(1000000000);
  scanf("%d %d", &al, &ar);
  scanf("%d %d", &bl, &br);
  scanf("%d", &k);
  printf("%.12lf\n", solve());
  return 0;
}