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;
#pragma comment(linker, "/STACK:255000000")
bool firstout = 1;
template <class T>
void minn(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
void maxx(T &a, T b) {
  if (a < b) a = b;
}
int n, m;
int A[10123];
int calc(int a, int aa, int l, int r) {
  if (l > aa || r < a || l > r) return 0.;
  maxx(l, a);
  minn(r, aa);
  return (r - l + 1);
}
int main() {
  int i, j, k;
  char c;
  int a, aa, b, bb, d;
  n = 0;
  for (k = (1); k < (10); ++k)
    for (i = (0); i < (1 << k); ++i) {
      d = 0;
      int dd = 1;
      for (j = (0); j < (k); ++j) {
        if ((i >> j) & 1)
          d += 7 * dd;
        else
          d += 4 * dd;
        dd *= 10;
      }
      A[n++] = d;
    }
  int ts;
  for (ts = 1; scanf("%"
                     "d",
                     &(a)) > 0;
       ++ts) {
    scanf(
        "%"
        "d",
        &(aa));
    scanf(
        "%"
        "d",
        &(b));
    scanf(
        "%"
        "d",
        &(bb));
    scanf(
        "%"
        "d",
        &(m));
    long long res = 0;
    for (i = (0); i < (n - m + 1); ++i) {
      int l, r;
      if (i == 0)
        l = 1;
      else
        l = A[i - 1] + 1;
      if (i + m == n)
        r = 1012345678;
      else
        r = A[i + m] - 1;
      res += (long long)calc(a, aa, l, A[i]) * calc(b, bb, A[i + m - 1], r);
      res += (long long)calc(b, bb, l, A[i]) * calc(a, aa, A[i + m - 1], r);
      if (m == 1 && a <= A[i] && A[i] <= aa && b <= A[i] && A[i] <= bb) --res;
    }
    printf(
        "%"
        ".10lf",
        ((double)res / (aa - a + 1) / (bb - b + 1)));
    printf("\n"), firstout = 1;
  }
  return 0;
}