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;
static const int INF = 500000000;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
int luck[2005], m;
int L1, R1, L2, R2, k;
int BIL = 1000000000;
void getluck(long long int a) {
  if (a > 0 && a <= BIL) luck[m++] = a;
  if (a > BIL) return;
  getluck(a * 10 + 4);
  getluck(a * 10 + 7);
}
int main() {
  luck[m++] = -10;
  luck[m++] = BIL + 10;
  getluck(0);
  sort(luck, luck + m);
  cin >> L1 >> R1 >> L2 >> R2 >> k;
  long long int res = 0;
  for (int i = 1; i < m - k; ++i) {
    res += max(min(luck[i], R1) - max(luck[i - 1] + 1, L1) + 1, 0) *
           (long long int)max(
               0, min(luck[i + k] - 1, R2) - max(luck[i + k - 1], L2) + 1);
    res += max(min(luck[i], R2) - max(luck[i - 1] + 1, L2) + 1, 0) *
           (long long int)max(
               0, min(luck[i + k] - 1, R1) - max(luck[i + k - 1], L1) + 1);
  }
  if (k == 1) {
    for (int i = 1; i < m - 1; ++i)
      if (L1 <= luck[i] && luck[i] <= R1 && L2 <= luck[i] && luck[i] <= R2)
        --res;
  }
  double ans = res / (double)((R1 - L1 + 1) * (long long int)(R2 - L2 + 1));
  printf("%.10f\n", ans);
  return 0;
}