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 gen(int n, long long x) {
  lucky.push_back(x);
  if (n > 0) {
    gen(n - 1, x * 10 + 4);
    gen(n - 1, x * 10 + 7);
  }
}
void init() {
  gen(10, 0);
  sort(lucky.begin(), lucky.end());
}
long long gao(long long l1, long long r1, long long l2, long long r2) {
  return max(0LL, min(r1, r2) - max(l1, l2) + 1);
}
int main() {
  long long l1, r1, l2, r2, a, b;
  int n;
  init();
  cin >> l1 >> r1 >> l2 >> r2 >> n;
  a = 0;
  b = (r1 - l1 + 1) * (r2 - l2 + 1);
  for (int i = n; i < (int)lucky.size() - 1; i++) {
    a += gao(l1, r1, lucky[i - n] + 1, lucky[i - n + 1]) *
         gao(l2, r2, lucky[i], lucky[i + 1] - 1);
    a += gao(l2, r2, lucky[i - n] + 1, lucky[i - n + 1]) *
         gao(l1, r1, lucky[i], lucky[i + 1] - 1);
    if (n == 1)
      a -= gao(l1, r1, lucky[i], lucky[i]) * gao(l2, r2, lucky[i], lucky[i]);
  }
  printf("%.12lf\n", 1.0 * a / b);
  return 0;
}