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;
long long min(long long a, long long b) { return (a < b ? a : b); }
long long max(long long a, long long b) { return (a > b ? a : b); }
long long lk[2222];
long long luc(int x) {
  int ans;
  if (x == 0) return 0;
  if (lk[x] == -1) {
    if (x % 2 == 0)
      lk[x] = luc(x / 2 - 1) * 10 + 7;
    else
      lk[x] = luc(x / 2) * 10 + 4;
  }
  return lk[x];
}
int main() {
  int a, b, c, d, k;
  scanf("%d %d %d %d %d", &a, &c, &b, &d, &k);
  memset(lk, -1, sizeof(lk));
  if (a > b || (a == b && c > d)) {
    a ^= b;
    c ^= d;
    b ^= a;
    d ^= c;
    a ^= b;
    c ^= d;
  }
  int l = 1, r = 2000;
  while (l < r) {
    int m = (l + r) / 2;
    if (luc(m) == a) {
      l = r = m;
      continue;
    }
    if (luc(m) < a) {
      l = m + 1;
    } else
      r = m;
  }
  double d1, d2;
  d2 = min(c - a + 1, d - b + 1);
  d1 = max(c - a + 1, d - b + 1);
  int ant = a;
  double prob = 0.0;
  for (; l + k - 1 < 2222 && (luc(l + k - 1) <= d || luc(l + k - 1) < c); l++) {
    long long xx, yy;
    xx = min(c, luc(l)) - max(a, luc(l - 1) + 1) + 1;
    yy = min(d, luc(l + k) - 1) - max(luc(l + k - 1), b) + 1;
    if (xx > 0 && yy > 0) {
      prob += ((double)(xx * yy)) / d1;
    }
    xx = min(d, luc(l)) - max(b, luc(l - 1) + 1) + 1;
    yy = min(c, luc(l + k) - 1) - max(luc(l + k - 1), a) + 1;
    if (xx > 0 && yy > 0) {
      prob += ((double)(xx * yy)) / d1;
    }
    if (k == 1 && luc(l) >= a && luc(l) <= c && luc(l) >= b && luc(l) <= d)
      prob -= ((double)1) / d1;
  }
  prob /= d2;
  printf("%.11lf\n", prob);
  return 0;
}